NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Updated on Aug 4, 2020 - 9:52 p.m. IST by Ravindra Pindel

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables:- It is an important chapter of algebra where you will learn to solve the linear equation having two variables. In NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables, you will be getting the entire coverage of the solutions including the optional exercise. Many real-life situations can be formulated using Mathematical equations. For example, consider the statement "cost of 1 Kg Apple and 2Kg orange is 120 and the cost of 3 Kg Apple and 1 Kg orange is 210". The statement can be formulated using the Mathematical equation, for this consider the cost of Apple as x and that of orange as y. Then we can write two equations as x+2y=120 and 3x+y=210. NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables, introduces the multiple methods to solve the two linear equations.

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The problems related to this chapter are very interesting and are important in the board exam and also for many other competitive exams. NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables will help you in boosting your preparation for all type of examinations. Here you will get NCERT Solutions from class 6 to 12 for science and maths. Here you will get NCERT solutions for class 10 also.

Types of questions asked from class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Solving a linear equation with two variables.
Representation of linear equation in a graph.
Solutions of linear equations using the graph.
Algebraic interpretation of linear equations.
Formation of linear equations using statements.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.1

Q1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Answer:

Let x be the age of Aftab and y be the age of his daughter

Now, According to the question,

$(x-7)=7\times(y-7)$

$\Rightarrow x-7=7y-49$

$\Rightarrow x-7y=-42.....(1)$

Also,

$(x+3)=3(y+3)$

$\Rightarrow x+3=3y+9$

$\Rightarrow x-3y=6.....(2)$

Now, let's represent both equations graphically,

From (1), we get

$y=\frac{x+42}{7}$

So, Putting different values of x we get corresponding values of y

X	0	7	-7
Y	6	7	5

And From (2) we get,

$y=\frac{x-6}{3}$

So, Putting different values of x we get corresponding values of y

X	0	3	6
Y	-2	-1	0

GRAPH:

Graph X-Y

Q2 The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Answer:

Let the price of one Bat be x and the price of one ball be y,

Now, According to the question,

$3x+6y=3900.....(1)$

$x+3y=1300.....(2)$

From(1) we have

$y=\frac{3900-3x}{6}$

By putting different values of x, we get different corresponding values of y.So

X	100	300	-100
Y	600	500	700

Now, From (2), we have,

$y=\frac{1300-x}{3}$

By putting different values of x, we get different corresponding values of y.So

X	100	400	-200
Y	400	300	500

GRAPH:

Q3 The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Answer:

Let, x be the cost of 1kg apple and y be the cost of 1kg grapes.

Now, According to the question,

On a day:

$2x+y=160.....(1)$

After One Month:

$4x+2y=300.....(2)$

Now, From (1) we have

$y=160-2x$

Putting different values of x we get corresponding values of y, so,'

X	80	60	50
Y	0	40	60

And From (2) we have,

$y=\frac{300-4x}{2}=150-2x$

Putting different values of x we get corresponding values of y, so,

X	50	60	70
Y	50	30	10

Graph:

graph

NCERT solutions for class 10 maths Chapter 3 Pair of Linear Equations in two variables Excercise: 3.2

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Answer:

Let the number of boys is x and the number of girls is y.

Now, According to the question,

Total number of students in the class = 10, i.e.

$\Rightarrow x+y=10.....(1)$

And

the number of girls is 4 more than the number of boys,i.e.

$x=y+4$

$\Rightarrow x-y=4..........(2)$

Different points (x, y) for equation (1)

X	5	6	4
Y	5	4	6

Different points (x,y) satisfying (2)

X	5	6	7
y	1	2	3

Graph,

As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3 which means the number of boys in the class is 7 and the number of girls in the class is 3.

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer:

Let x be the price of 1 pencil and y be the price of 1 pen,

Now, According to the question

$5x+7y=50......(1)$

And

$7x+5y=46......(2)$

Now, the points (x,y), that satisfies the equation (1) are

X	3	-4	10
Y	5	10	0

And, the points(x,y) that satisfies the equation (2) are

X	3	8	-2
Y	5	-2	12

The Graph,

Graph

As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.

Q2 On comparing the ratios $\frac{a_1}{a_2}$ , $\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) $\\5x - 4y + 8 = 0\\ 7x + 6y - 9 = 0$

Answer:

Give, Equations,

$\\5x - 4y + 8 = 0\\ 7x + 6y - 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{5}{7},\:\frac{b_1}{b_2}=\frac{-4}{6}\:and\:\frac{c_1}{c_2}=\frac{8}{-9}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means that both lines intersect at exactly one point.

Answer:

Given, Equations,

$\\9x + 3y + 12 = 0\\ 18x + 6y + 24 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\:and \\\:\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means that both lines are coincident.

Answer:

Give, Equations,

$\\6x - 3y + 10 = 0\\ 2x - y+ 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{6}{2}=3,\:\frac{b_1}{b_2}=\frac{-3}{-1}=3\:and\:\frac{c_1}{c_2}=\frac{10}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means that both lines are parallel to each other.

Q2 On comparing the ratios $\frac{a_1}{a_2}$ , $\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (i) $3x + 2y = 5;\qquad 2x - 3y = 7$

Answer:

Give, Equations,

$\\3x + 2y = 5;\qquad\\ 2x - 3y = 7$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{2},\:\frac{b_1}{b_2}=\frac{2}{-3}\:and\:\frac{c_1}{c_2}=\frac{5}{7}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Q3 On comparing the ratios $\frac{a_1}{a_2}$ , $\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (ii) $2x - 3y = 8;\qquad 4x - 6y = 9$

Answer:

Given, Equations,

$\\2x - 3y = 8;\qquad \\4x - 6y = 9$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{8}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Answer:

Given, Equations,

$\\\frac{3}{2}x + \frac{5}{3}y = 7;\qquad\\ \\ 9x -10y = 14$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{18}=\frac{1}{6},\\\:\frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-30}=-\frac{1}{6}\:and\\\:\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Answer:

Given, Equations,

$5x - 3y = 11;\qquad \\-10x + 6y =-22$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{11}{-22}=-\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Q3 On comparing the ratios $\frac{a_1}{a_2}$ , $\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$ , find out whether the following pair of linear equations are consistent, or inconsistent (v) $\frac{4}{3}x + 2y = 8; \qquad 2x + 3y = 12$

Answer:

Given, Equations,

$\\\frac{4}{3}x + 2y = 8; \qquad\\\\ 2x + 3y = 12$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{4/3}{2}=\frac{4}{6}=\frac{2}{3},\\\:\frac{b_1}{b_2}=\frac{2}{3}\:\:and\\\:\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) $x + y = 5 \qquad 2x + 2 y = 10$

Answer:

Given, Equations,

$\\x + y = 5 \qquad\\ 2x + 2 y = 10$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

The points (x,y) which satisfies in both equations are

X	1	3	5
Y	4	2	0

Graph

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (ii) $x - y = 8,\qquad 3x - 3y = 16$

Answer:

Given, Equations,

$\\x - y = 8,\qquad\\ 3x - 3y = 16$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{3},\\\:\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\:and\\\:\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (iii) $2x + y - 6 =0, \qquad 4x - 2 y - 4 = 0$

Answer:

Given, Equations,

$\\2x + y - 6 =0, \qquad \\4x - 2 y - 4 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Now The points(x, y) satisfying the equation are,

X	0	2	3
Y	6	2	0

And The points(x,y) satisfying the equation $\\4x - 2 y - 4 = 0$ are,

X	0	1	2
Y	-2	0	2

GRAPH:

As we can see both lines intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

Q4 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (iv) $2x - 2y - 2 =0, \qquad 4x - 4y -5 = 0$

Answer:

Given, Equations,

$\\2x - 2y - 2 =0, \qquad\\ 4x - 4y -5 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Q5 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let $l$ be the length of the rectangular garden and $b$ be the width.

Now, According to the question, the length is 4 m more than its width.i.e.

$l=b+4$

$l-b=4....(1)$

Also Given Half Parameter of the rectangle = 36 i.e.

$l+b=36....(2)$

Now, as we have two equations, on adding both equations, we get,

$l+b+l-b=4+36$

$\Rightarrow 2l=40$

$\Rightarrow l=20$

Putting this in equation (1),

$\Rightarrow 20-b=4$

$\Rightarrow b=20-4$

$\Rightarrow b=16$

Hence Length and width of the rectangle are 20m and 16 respectively.

Q6 Given the linear equation $2x + 3y -8 =0$ , write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines

Answer:

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the intersection of lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is ,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

So Any line with this condition can be $4x+3y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{3}=1$

$\frac{1}{2}\neq1$ the line satisfies the given condition.

Q6 Given the linear equation $2x + 3y -8 =0$ , write another linear equation in two variables such that the geometrical representation of the pair so formed is (ii) parallel lines

Answer:

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , for being parallel is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So Any line with this condition can be $4x+6y-8=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-8}=1$

$\frac{1}{2}=\frac{1}{2}\neq1$ the line satisfies the given condition.

Q6 Given the linear equation $2x + 3y -8 =0$ , write another linear equation in two variables such that the geometrical representation of the pair so formed is: (iii) coincident lines

Answer:

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the coincidence of the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So any line with this condition can be $4x+6y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}$

$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$ the line satisfies the given condition.

Q7 Draw the graphs of the equations $x - y + 1=0$ and $3x +2 y - 12=0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer:

Given, two equations,

$x - y + 1=0.........(1)$

And

$3x +2 y - 12=0.........(2)$

The points (x,y) satisfying (1) are

X	0	3	6
Y	1	4	7

And The points(x,y) satisfying (2) are,

X	0	2	4
Y	6	3	0

GRAPH:

24151

As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green color.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.3

Q1 Solve the following pair of linear equations by the substitution method. (i) $\\x + y = 14\\ x - y = 4$

Answer:

Given, two equations,

$\\x + y = 14.......(1)\\ x - y = 4........(2)$

Now, from (1), we have

$y=14-x........(3)$

Substituting this in (2), we get

$x-(14-x)=4$

$\Rightarrow x-14+x=4$

$\Rightarrow 2x=4+14=18$

$\Rightarrow x=9$

Substituting this value of x in (3)

$\Rightarrow y=14-x=14-9=5$

Hence, Solution of the given equations is x = 9 and y = 5.

Q1 Solve the following pair of linear equations by the substitution method (ii) $\\s - t = 3\\ \frac{s}{3} + \frac{t}{2} = 6$

Answer:

Given, two equations,

$\\s - t = 3.........(1)\\ \frac{s}{3} + \frac{t}{2} = 6.......(2)$

Now, from (1), we have

$s=t+3........(3)$

Substituting this in (2), we get

$\frac{t+3}{3}+\frac{t}{2}=6$

$\Rightarrow \frac{2t+6+3t}{6}=6$

$\Rightarrow 5t+6=36$

$\Rightarrow 5t=30$

$\Rightarrow t=6$

Substituting this value of t in (3)

$\Rightarrow s=t+3 = 6+3=9$

Hence, Solution of the given equations is s = 9 and t = 6.

Q1 Solve the following pair of linear equations by the substitution method. (iii) $\\ 3 x - y = 3\\ 9x - 3y = 9$

Answer:

Given, two equations,

$\\ 3 x - y = 3......(1)\\ 9x - 3y = 9.....(2)$

Now, from (1), we have

$y=3x-3........(3)$

Substituting this in (2), we get

$9x-3(3x-3)=9$

$\Rightarrow 9x-9x+9=9$

$\Rightarrow 9=9$

This is always true, and hence this pair of the equation has infinite solutions.

As we have

$y=3x-3$ ,

One of many possible solutions is $x=1,\:and\:y=0$ .

Q1 Solve the following pair of linear equations by the substitution method. (iv) $\\0.2x + 0.3y = 1.3\\ 0.4x + 0.5y = 2.3$

Answer:

Given, two equations,

$\\0.2x + 0.3y = 1.3\\ 0.4x + 0.5y = 2.3$

Now, from (1), we have

$y=\frac{1.3-0.2x}{0.3}........(3)$

Substituting this in (2), we get

$0.4x+0.5\left(\frac{1.3-0.2x}{0.3}\right)=2.3$

$\Rightarrow 0.12x+0.65-0.1x=0.69$

$\Rightarrow 0.02x=0.69-0.65=0.04$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{1.3-0.2x}{0.3} \right )=\left ( \frac{1.3-0.4}{0.3} \right )=\frac{0.9}{0.3}=3$

Hence, Solution of the given equations is,

$x=2\:and\:y=3$ .

Q1 Solve the following pair of linear equations by the substitution method. (v) $\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Answer:

Given, two equations,

$\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Now, from (1), we have

$y=-\frac{\sqrt{2}x}{\sqrt{3}}........(3)$

Substituting this in (2), we get

$\sqrt{3}x-\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )=0$

$\Rightarrow \sqrt{3}x=\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )$

$\Rightarrow \3x=-4x$

$\Rightarrow7x=0$

$\Rightarrow x=0$

Substituting this value of x in (3)

$\Rightarrow y=-\frac{\sqrt{2}x}{\sqrt{3}}=0$

Hence, Solution of the given equations is,

$x=0,\:and \:y=0$ .

Q1 Solve the following pair of linear equations by the substitution method. (vi) $\\\frac{3x}{2} - \frac{5y}{3}= - 2\\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Answer:

Given,

$\\\frac{3x}{2} - \frac{5y}{3}= - 2.........(1)\\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}.............(2)$

From (1) we have,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )........(3)$

Putting this in (2) we get,

$\frac{1}{3}\times\frac{2}{3}\left ( \frac{5y}{3}-2 \right )+\frac{y}{2}=\frac{13}{6}$

$\frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}$

$\frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}$

$\frac{47y}{54}=\frac{13}{6}+\frac{4}{9}$

$\frac{47y}{54}=\frac{117}{54}+\frac{24}{54}$

$47y=117+24$

$47y=141$

$y=\frac{141}{47}$

$y=3$

putting this value in (3) we get,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )$

$x=\frac{2}{3}\left ( \frac{5\times3}{3}-2 \right )$

$x=\frac{2}{3}\left (5-2 \right )$

$x=\frac{2}{3}\times 3$

$x=2$

Hence $x=2\:\:and\:\:y=3.$

Q2 Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of ‘ $m$ ’ for which $y = mx + 3$ .

Answer:

Given, two equations,

$2x + 3y = 11......(1)$

$2x - 4y = -24.......(2)$

Now, from (1), we have

$y=\frac{11-2x}{3}........(3)$

Substituting this in (2), we get

$2x-4\left ( \frac{11-2x}{3} \right )=-24$

$\Rightarrow 6x-44+8x=-72$

$\Rightarrow 14x=44-72$

$\Rightarrow 14x=-28$

$\Rightarrow x=-2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{11-2x}{3} \right )=\frac{11-2\times(-2)}{3}=\frac{15}{3}=5$

Hence, Solution of the given equations is,

$x=-2,\:and\:y=5.$

Now,

As it satisfies $y=mx+3$ ,

$\Rightarrow 5=m(-2)+3$

$\Rightarrow 2m=3-5$

$\Rightarrow 2m=-2$

$\Rightarrow m=-1$

Hence Value of m is -1.

Q3 Form the pair of linear equations for the following problem and find their solution by substitution method.

(i) The difference between the two numbers is 26 and one number is three times the other. Find them.

Answer:

Let two numbers be x and y and let the bigger number is y.

Now, According to the question,

$y-x=26......(1)$

And

$y=3x......(2)$

Now, the substituting value of y from (2) in (1) we get,

$3x-x=26$

$\Rightarrow 2x=26$

$\Rightarrow x=13$

Substituting this in (2)

$\Rightarrow y=3x=3(13)=39$

Hence the two numbers are 13 and 39.

Q3 Form the pair of linear equations for the following problem and find their solution by substitution method (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and smaller angle be y

Now, As we know the sum of supplementary angles is 180. so,

$x+y=180.......(1)$

Also given in the question,

$x-y=18.......(2)$

Now, From (2) we have,

$y=x-18.......(3)$

Substituting this value in (1)

$x+x-18=180$

$\Rightarrow 2x=180+18$

$\Rightarrow 2x=198$

$\Rightarrow x=99$

Now, Substituting this value of x in (3), we get

$\Rightarrow y=x-18=99-18=81$

Hence the two supplementary angles are

$99^0\:and\:81^0.$

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer:

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, According to the question,

$7x+6y=3800......(1)$

$3x+5y=1750......(2)$

Now, From (1) we have

$y=\frac{3800-7x}{6}........(3)$

Substituting this value of y in (2)

$3x+5\left ( \frac{3800-7x}{6} \right )=1750$

$\Rightarrow 18x+19000-35x=1750\times6$

$\Rightarrow -17x=10500-19000$

$\Rightarrow -17x=-8500$

$\Rightarrow x=\frac{8500}{17}$

$\Rightarrow x=500$

Now, Substituting this value of x in (3)

$y=\frac{3800-7x}{6}=\frac{3800-7\times500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$

Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Answer:

Let the fixed charge is x and the per km charge is y.

Now According to the question

$x+10y=105.......(1)$

And

$x+15y=155.......(2)$

Now, From (1) we have,

$x=105-10y........(3)$

Substituting this value of x in (2), we have

$105-10y+15y=155$

$\Rightarrow 5y=155-105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, Substituting this value in (3)

$x=105-10y=105-10(10)=105-100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

$\Rightarrow x+25y=5+25(10)=5+250=255$

Hence fair for 25km is 255 Rs.

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes $\frac{9}{11 }$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5 }{6}$ . Find the fraction.

Answer:

Let the numerator of the fraction be x and denominator of the fraction is y

Now According to the question,

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x-9y=-4...........(1)$

Also,

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x-5y=-3...........(2)$

Now, From (1) we have

$y=\frac{11x+4}{9}.............(3)$

Substituting this value of y in (2)

$6x-5\left ( \frac{11x+4}{9} \right )=-3$

$\Rightarrow 54x-55x-20=-27$

$\Rightarrow -x=20-27$

$\Rightarrow x=7$

Substituting this value of x in (3)

$y=\frac{11x+4}{9}=\frac{11(7)+4}{9}=\frac{81}{9}=9$

Hence the required fraction is

$\frac{x}{y}=\frac{7}{9}.$

Q3 Form the pair of linear equations for the following problems and find their solution by substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

Let x be the age of Jacob and y be the age of Jacob's son,

Now, According to the question

$x+5=3(y+5)$

$\Rightarrow x+5=3y+15$

$\Rightarrow x-3y=10..........(1)$

Also,

$x-5=7(y-5)$

$\Rightarrow x-5=7y-35$

$\Rightarrow x-7y=-30.........(2)$

Now,

From (1) we have,

$x=10+3y...........(3)$

Substituting this value of x in (2)

$10+3y-7y=-30$

$\Rightarrow -4y=-30-10$

$\Rightarrow 4y=40$

$\Rightarrow y=10$

Substituting this value of y in (3),

$x=10+3y=10+3(10)=10+30=40$

Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.4

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :

(i) $x + y =5 \ \textup{and} \ 2x - 3y = 4$

Answer:

Elimination Method:

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, multiplying (1) by 3 we, get

$\\3x +3 y =15............(3)$

Now, Adding (2) and (3), we get

$\\2x-3y+3x +3 y =4+15$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value in (1) we, get

$\frac{19}{5}+y=5$

$\Rightarrow y=5-\frac{19}{5}$

$\Rightarrow y=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

Substitution method :

Given, equations

$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$

Now, from (1) we have,

$y=5-x.......(3)$

substituting this value in (2)

$2x-3(5-x)=4$

$\Rightarrow 2x-15+3x=4$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value of x in (3)

$\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}$

Hence,

$x=\frac{19}{5}\:and\:y=\frac{6}{5}$

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :

(ii) $3x + 4 y = 10 \ \textup{and} \ 2x - 2y = 2$

Answer:

Elimination Method:

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, multiplying (2) by 2 we, get

$\\4x -4 y =4............(3)$

Now, Adding (1) and (3), we get

$\\3x+4y+4x -4 y =10+4$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2(2)-2y=2$

$\Rightarrow 2y=2$

$\Rightarrow y=1$

Hence,

$x=2\:and\:y=1$

Substitution method :

Given, equations

$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$

Now, from (2) we have,

$y=\frac{2x-2}{2}=x-1.......(3)$

substituting this value in (1)

$3x+4(x-1)=10$

$\Rightarrow 3x+4x-4=10$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=x-1=2-1=1$

Hence,

$x=2\:and\:y=1$

Q1 Solve the following pair of linear equations by the elimination method and the substitution method: (iii) $3x - 5y -4 = 0\ \textup{and} \ 9x = 2y + 7$

Answer:

Elimination Method:

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, multiplying (1) by 3 we, get

$\\9x -15 y -12=0............(3)$

Now, Subtracting (3) from (2), we get

$9x-2y-7-9x+15y+12=0$

$\Rightarrow 13y+5=0$

$\Rightarrow y=\frac{-5}{13}$

Putting this value in (1) we, get

$3x-5(\frac{-5}{13})-4=0$

$\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3x=\frac{27}{13}$

$\Rightarrow x=\frac{9}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Substitution method :

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, from (2) we have,

$y=\frac{9x-7}{2}.......(3)$

substituting this value in (1)

$3x-5\left(\frac{9x-7}{2} \right )-4=0$

$\Rightarrow 6x-45x+35-8=0$

$\Rightarrow -39x+27=0$

$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$

Substituting this value of x in (3)

$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Q1 Solve the following pair of linear equations by the elimination method and the substitution method :(iv) $\frac{x}{2} + \frac{2y}{3} = -1\ \textup{and} \ x - \frac{y}{3} = 3$

Answer:

Elimination Method:

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, multiplying (2) by 2 we, get

$\\2x - \frac{2y}{3} =6............(3)$

Now, Adding (1) and (3), we get

$\\\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Putting this value in (2) we, get

$2-\frac{y}{3}=3$

$\Rightarrow \frac{y}{3}=-1$

$\Rightarrow y=-3$

Hence,

$x=2\:and\:y=-3$

Substitution method :

Given, equations

$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$

Now, from (2) we have,

$y=3(x-3)......(3)$

substituting this value in (1)

$\frac{x}{2}+\frac{2(3(x-3))}{3}=-1$

$\Rightarrow \frac{x}{2}+2x-6=-1$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=3(x-3)=3(2-1)=-3$

Hence,

$x=2\:and\:y=-3$

Q2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2 }$ if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

$\frac{x+1}{y-1}=1$

$\Rightarrow x+1=y-1$

$\Rightarrow x-y=-2.........(1)$

Also,

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x=y+1$

$\Rightarrow 2x-y=1..........(2)$

Now, Subtracting (1) from (2) we get

$x=3$

Putting this value in (1)

$3-y=-2$

$\Rightarrow y=5$

Hence

$x=3\:and\:y=5$

And the fraction is

$\frac{3}{5}$

Q2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the age of Nuri be x and age of Sonu be y.

Now, According to the question

$x-5=3(y-5)$

$\Rightarrow x-5=3y-15$

$\Rightarrow x-3y=-10.........(1)$

Also,

$x+10=2 (y+10)$

$\Rightarrow x+10=2y+20$

$\Rightarrow x-2y=10........(2)$

Now, Subtracting (1) from (2), we get

$y=20$

putting this value in (2)

$x-2(20)=10$

$\Rightarrow x=50$

Hence the age of Nuri is 50 and the age of Nuri is 20.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the unit digit of the number be x and 10's digit be y.

Now, According to the question,

$x+y=9.......(1)$ 24323

Also

$9(10y+x)=2(10x+y)$

$\Rightarrow 90y+9x=20x+2y$

$\Rightarrow 88y-11x=0$

$\Rightarrow 8y-x=0.........(2)$

Now adding (1) and (2) we get,

$\Rightarrow 9y=9$

$\Rightarrow y=1$

now putting this value in (1)

$x+1=9$

$\Rightarrow x=8$

Hence the number is 18.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer:

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, According to the question,

$x+y=25..........(1)$

And

$50x+100y=2000$

$\Rightarrow x+2y=40.............(2)$

Now, Subtracting(1) from (2), we get

$y=15$

Putting this value in (1).

$x+15=25$

$\Rightarrow x=10$

Hence Meena received 10 50 Rs notes and 15 100 Rs notes.

Q2 Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

Let fixed charge be x and per day charge is y.

Now, According to the question,

$x+4y=27...........(1)$

And

$x+2y=21...........(2)$

Now, Subtracting (2) from (1). we get,

$4y-2y=27-21$

$\Rightarrow 2y=6$

$\Rightarrow y=3$

Putting this in (1)

$x+4(3)=27$

$\Rightarrow x=27-12=15$

Hence the fixed charge is 15 Rs and per day charge is 3 Rs.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.5

Q1 Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using a cross multiplication method.

(i) $\\x - 3y -3 = 0\\ 3x - 9y -2 = 0$

Answer:

Given, two equations,

$\\x - 3y -3 = 0.........(1)\\ 3x - 9y -2 = 0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}$

$\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has no solution.

(ii) $\\2x + y = 5 \\ 3x + 2y = 8$

Answer:

Given, two equations,

$\\2x + y = 5.........(1) \\ 3x + 2y = 8..........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{2}{3}$

$\frac{b_1}{b_2}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{5}{8}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(1)(-8)-(2)(-5)}=\frac{y}{(-5)(3)-(-8)(2)}=\frac{1}{(2)(2)-(3)(1)}$

$\frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}$

$\frac{x}{2}=\frac{y}{1}=\frac{1}{1}$

$x=2,\:and\:y=1$

(iii) $\\3x -5 y = 20\\ 6x - 10y = 40$

Answer:

Given the equations,

$\\3x -5 y = 20..........(1)\\ 6x - 10y = 40........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, the pair of equations has infinitely many solutions.

(iv) $\\x - 3y -7 = 0\\ 3x -3y -15 =0$

Answer:

Given the equations,

$\\x - 3y -7 = 0.........(1)\\ 3x -3y -15 =0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-3}=1$

$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-3)-(3)(-3)}$

$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$

$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$x=\frac{24}{6}=4,\:and\:y=-1$

Q2 (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? $\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$

Answer:

Given equations,

$\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have an infinite solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$

From here we get,

$\frac{2}{a-b}=\frac{3}{a+b}$

$\Rightarrow 2(a+b)=3(a-b)$

$\Rightarrow 2a+2b=3a-3b$

$\Rightarrow a-5b=0.........(1)$

Also,

$\frac{2}{a-b}=\frac{7}{3a+b-2}$

$\Rightarrow 2(3a+b-2)=7(a-b)$

$\Rightarrow 6a+2b-4=7a-7b$

$\Rightarrow a-9b+4=0...........(2)$

Now, Subtracting (2) from (1) we get

$\Rightarrow 4b-4=0$

$\Rightarrow b=1$

Substituting this value in (1)

$\Rightarrow a-5(1)=0$

$\Rightarrow a=5$

Hence, $a=5\:and\:b=1$ .

Q2 (ii) For which value of k will the following pair of linear equations have no solution? $\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

Answer:

Given, the equations,

$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have no solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{1}{2k+1}$

From here we get,

$\frac{3}{2k-1}=\frac{1}{k-1}$

$\Rightarrow 3(k-1)=2k-1$

$\Rightarrow 3k-3=2k-1$

$\Rightarrow 3k-2k=3-1$

$\Rightarrow k=2$

Hence, the value of K is 2.

Q3 Solve the following pair of linear equations by the substitution and cross-multiplication methods :
$\\8x + 5y = 9 \\3x + 2y = 4$

Answer:

Given the equations

$\\8x + 5y = 9........(1) \\3x + 2y = 4........(2)$

By Substitution Method,

From (1) we have

$y=\frac{9-8x}{5}.........(3)$

Substituting this in (2),

$3x+2\left ( \frac{9-8x}{5} \right )=4$

$\Rightarrow 15x+18-16x=20$

$\Rightarrow -x=20-18$

$\Rightarrow x=-2$

Substituting this in (3)

$y=\frac{9-8x}{5}=\frac{9-8(-2)}{5}=\frac{25}{5}=5$

Hence $x=-2\:and\:y=5$ .

By Cross Multiplication Method

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(5)(-4)-(2)(-9)}=\frac{y}{(3)(-9)-(8)(-4)}=\frac{1}{(8)(2)-(3)(5)}$

$\frac{x}{-20+18}=\frac{y}{32-27}=\frac{1}{16-15}$

$\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$

$x=-2,\:and\:y=5$

Q4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Answer:

Let the fixed charge be x and the cost of food per day is y,

Now, According to the question

$x+20y=1000.........(1)$

Also

$x+26y=1180.........(2)$

Now subtracting (1) from (2),

$x+26y-x-20y=1180-100$

$\Rightarrow 6y=180$

$\Rightarrow y=30$

Putting this value in (1)

$x+20(30)=1000$

$\Rightarrow x=1000-600$

$\Rightarrow x=400$

Hence, the Fixed charge is Rs 400 and the cost of food per day is Rs 30.

Q4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (ii) A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

Answer:

Let numerator of a fraction be x and the denominator is y.

Now, According to the question,

$\frac{x-1}{y}=\frac{1}{3}$

$\Rightarrow 3(x-1)=y$

$\Rightarrow 3x-3=y$

$\Rightarrow 3x-y=3........(1)$

Also,

$\frac{x}{y+8}=\frac{1}{4}$

$\Rightarrow 4x=y+8$

$\Rightarrow 4x-y=8.........(2)$

Now, Subtracting (1) from (2) we get,

$4x-3x=8-3$

$\Rightarrow x=5$

Putting this value in (2) we get,

$4(5)-y=8$

$\Rightarrow y=20-8$

$\Rightarrow y=12$

Hence, the fraction is

$\frac{x}{y}=\frac{5}{12}$ .

Q4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Answer:

Let the number of right answer and wrong answer be x and y respectively

Now, According to the question,

$3x-y=40..........(1)$

And

$\\4x-2y=50\\\Rightarrow 2x-y=25..........(2)$

Now, subtracting (2) from (1) we get,

$x=40-25$

$x=15$

Putting this value in (1)

$3(15)-y=40$

$\Rightarrow y=45-40$

$\Rightarrow y=5$

Hence the total number of question is $x+y=15+5=20.$

Q4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Answer:

Let the speed of the first car is x and the speed of the second car is y.

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction= x - y

the relative speed when they are going in the opposite direction= x + y

The given relative distance between them = 100 km.

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$5\times(x-y)=100$

$\Rightarrow 5x-5y=100$

$\Rightarrow x-y=20.........(1)$

Also,

$1(x+y)=100$

$\Rightarrow x+y=100........(2)$

Now Adding (1) and (2) we get

$2x=120$

$\Rightarrow x=60$

putting this in (1)

$60-y=20$

$\Rightarrow y=60-20$

$\Rightarrow y=40$

Hence the speeds of the cars are 40 km/hour and 60 km/hour.

Q4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (v) The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer:

Let $l$ be the length of the rectangle and $b$ be the width,

Now, According to the question,

$(l-5)(b+3)=lb-9$

$\Rightarrow lb+3l-5b-15=lb-9$

$\Rightarrow 3l-5b-6=0..........(1)$

Also,

$(l+3)(b+2)=lb+67$

$\Rightarrow lb+2l+3b+6=lb+67$

$\Rightarrow 2l+3b-61=0..........(2)$

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-5)(-61)-(3)(-6)}=\frac{y}{(-6)(2)-(-61)(3)}=\frac{1}{(3)(3)-(2)(-5)}$

$\frac{x}{305+18}=\frac{y}{-12+183}=\frac{1}{9+10}$

$\frac{x}{323}=\frac{y}{171}=\frac{1}{19}$

$x=17,\:and\:y=9$

Hence the length and width of the rectangle are 17 units and 9 units respectively.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.6

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) $\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Answer:

Given Equations,

$\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow 3p+2q=12........(1)$

And

$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow 2p+3q=13..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(2)(-13)-(3)(-12)}=\frac{q}{(-12)(2)-(-13)(3)}=\frac{1}{(3)(3)-(2)(2)}$

$\frac{p}{-26+36}=\frac{q}{-24+39}=\frac{1}{9-4}$

$\frac{p}{10}=\frac{q}{15}=\frac{1}{5}$

$p=2,\:and\:q=3$

And Hence,

$x=\frac{1}{2}\:and\:y=\frac{1}{3}.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(ii) $\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Answer:

Given Equations,

$\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Let,

$\frac{1}{\sqrt{x}}=p\:and\:\frac{1}{\sqrt{y}}=q$

Now, our equation becomes

$2p+3q=2........(1)$

And

$4p-9q=-1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(3)(1)-(-9)(-2)}=\frac{q}{(-2)(4)-(1)(2)}=\frac{1}{(2)(-9)-(4)(3)}$

$\frac{p}{3-18}=\frac{q}{-8-2}=\frac{1}{-18-12}$

$\frac{p}{-15}=\frac{q}{-10}=\frac{1}{-30}$

$p=\frac{1}{2},\:and\:q=\frac{1}{3}$

So,

$p=\frac{1}{2}=\frac{1}{\sqrt{x}}\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{\sqrt{y}}\Rightarrow y=9$ .

And hence

$x=4\:and\:y=9.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(iii) $\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Answer:

Given Equations,

$\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Let,

$\frac{1}{x}=p\:and\:y=q$

Now, our equation becomes

$\Rightarrow 4p+3q=14........(1)$

And

$\Rightarrow 3p-4q=23..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(3)(-23)-(-4)(-14)}=\frac{q}{(-14)(3)-(-23)(4)}=\frac{1}{(4)(-4)-(3)(3)}$

$\frac{p}{-69-56}=\frac{q}{-42+92}=\frac{1}{-16-9}$

$\frac{p}{-125}=\frac{q}{50}=-\frac{1}{25}$

$p=5,\:and\:q=-2$

And Hence,

$x=\frac{1}{5}\:and\:y=-2.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(iv) $\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Answer:

Given Equations,

$\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Let,

$\frac{1}{x-1}=p\:and\:\frac{1}{y-2}=q$

Now, our equation becomes

$5p+q=2........(1)$

And

$6p-3q=1..........(2)$

Multiplying (1) by 3 we get

$15p+3q=6..........(3)$

Now, adding (2) and (3) we get

$21p=7$

$\Rightarrow p=\frac{1}{3}$

Putting this in (2)

$6\left ( \frac{1}{3} \right )-3q=1$

$\Rightarrow 3q=1$

$\Rightarrow q=\frac{1}{3}$

Now,

$p=\frac{1}{3}=\frac{1}{x-1}\Rightarrow x-1=3\Rightarrow x=4$

$q=\frac{1}{3}=\frac{1}{y-2}\Rightarrow y-2=3\Rightarrow x=5$

Hence,

$x=4,\:and\:y=5.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v)

$\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15$

Answer:

Given Equations,

$\\\frac{7x - 2y}{xy} = 5\\\\\Rightarrow\frac{7}{y} -\frac{2}{x}=5\\ \frac{8x + 7y}{xy} = 15\\\Rightarrow \frac{8}{y}+\frac{7}{x}=15$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$7q-2p=5........(1)$

And

$8q+7p=15..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(-2)(-15)-(7)(-5)}=\frac{p}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)}$

$\frac{q}{30+35}=\frac{p}{-40+105}=\frac{1}{49 +16}$

$\frac{q}{65}=\frac{p}{65}=\frac{1}{65}$

$p=1,\:and\:q=1$

And Hence,

$x=1\:and\:y=1.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(vi) $\\6x + 3y = 6xy\\ 2x + 4y = 5 xy$

Answer:

Given Equations,

$\\6x + 3y = 6xy\\\Rightarrow \frac{6x}{xy}+\frac{3y}{xy}=6\\\\\Rightarrow \frac{6}{y}+\frac{3}{x}=6\\and\\\ 2x + 4y = 5 xy\\\Rightarrow \frac{2x}{xy}+\frac{4y}{xy}=5\\\Rightarrow \frac{2}{y}+\frac{4}{x}=5$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$6q+3p=6........(1)$

And

$2q+4p=5..........(2)$

By Cross Multiplication method,

$\frac{q}{b_1c_2-b_2c_1}=\frac{p}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(3)(-5)-(-6)(4)}=\frac{p}{(6)(2)-(6)(-5)}=\frac{1}{(6)(4)-(3)(2)}$

$\frac{q}{-15+24}=\frac{p}{-12+30}=\frac{1}{24 -6}$

$\frac{q}{9}=\frac{p}{18}=\frac{1}{18}$

$q=\frac{1}{2}\:and\:p=1$

And Hence,

$x=1\:and\:y=2.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:
(vii) $\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2$

Answer:

Given Equations,

$\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2$

Let,

$\frac{1}{x+y}=p\:and\:\frac{1}{x-y}=q$

Now, our equation becomes

$10p+2q=4........(1)$

And

$15p-5q=-2..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(2)(2)-(-5)(-4)}=\frac{q}{(-4)(15)-(2)(10)}=\frac{1}{(10)(-5)-(15)(2)}$

$\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}$

$\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}$

$p=\frac{1}{5},\:and\:q=1$

Now,

$p=\frac{1}{5}=\frac{1}{x+y}$

$\Rightarrow x+y=5........(3)$

And,

$q=1=\frac{1}{x-y}$

$\Rightarrow x-y=1...........(4)$

Adding (3) and (4) we get,

$\Rightarrow 2x=6$

$\Rightarrow x=3$

Putting this value in (3) we get,

$3+y=5$

$\Rightarrow y=2$

And Hence,

$x=3\:and\:y=2.$

Q1 Solve the following pairs of equations by reducing them to a pair of linear equations:

(viii) $\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Answer:

Given Equations,

$\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Let,

$\frac{1}{3x+y}=p\:and\:\frac{1}{3x-y}=q$

Now, our equation becomes

$p+q=\frac{3}{4}.........(1)$

And

$\\\frac{p}{2}-\frac{q}{2}=\frac{-1}{8}\\\\p-q=\frac{-1}{4}..........(2)$

Now, Adding (1) and (2), we get

$2p=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow 2p=\frac{2}{4}$

$\Rightarrow p=\frac{1}{4}$

Putting this value in (1)

$\frac{1}{4}+q=\frac{3}{4}$

$\Rightarrow q=\frac{3}{4}-\frac{1}{4}$

$\Rightarrow q=\frac{2}{4}$

$\Rightarrow q=\frac{1}{2}$

Now,

$p=\frac{1}{4}=\frac{1}{3x+y}$

$\Rightarrow 3x+y=4...........(3)$

And

$q=\frac{1}{2}=\frac{1}{3x-y}$

$\Rightarrow 3x-y=2............(4)$

Now, Adding (3) and (4), we get

$6x=4+2$

$\Rightarrow 6x=6$

$\Rightarrow x=1$

Putting this value in (3),

$3(1)+y=4$

$\Rightarrow y=4-3$

$\Rightarrow y=1$

Hence,

$x=1,\:and\:y=1$

Q2 Formulate the following problems as a pair of equations and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Answer:

Let the speed of Ritu in still water be x and speed of current be y,

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction (downstream)= x +y

the relative speed when they are going in the opposite direction (upstream)= x - y

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$x+y=\frac{20}{2}$

$\Rightarrow x+y=10.........(1)$

And,

$x-y=\frac{4}{2}$

$\Rightarrow x-y=2...........(2)$

Now, Adding (1) and (2), we get

$2x=10+2$

$\Rightarrow 2x=12$

$\Rightarrow x=6$

Putting this in (2)

$6-y=2$

$\Rightarrow y=6-2$

$\Rightarrow y=4$

Hence,

$x=6\:and\:y=4.$

Hence Speed of Ritu in still water is 6 km/hour and speed of the current is 4 km/hour

Q2 Formulate the following problems as a pair of equations and hence find their solutions: (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Answer:

Let the number of days taken by woman and man be x and y respectively,

The proportion of Work done by a woman in a single day

$=\frac{1}{x }$

The proportion of Work done by a man in a single day

$=\frac{1}{y }$

Now, According to the question,

$4\left ( \frac{2}{x}+\frac{5}{y} \right )=1$

$\Rightarrow \left ( \frac{2}{x}+\frac{5}{y} \right )=\frac{1}{4}$

Also,

$3\left ( \frac{3}{x}+\frac{6}{y} \right )=1$

$\Rightarrow \left ( \frac{3}{x}+\frac{6}{y} \right )=\frac{1}{3}$

Let,

$\frac{1}{x}=p\:and\:\frac{1}{y}=q$

Now, our equation becomes

$2p+5q=\frac{1}{4}$

$8p+20q=1........(1)$

And

$3p+6p=\frac{1}{3}$

$\Rightarrow 9p+18p=1.............(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{p}{(20)(-1)-(18)(-1)}=\frac{q}{(-1)(9)-(8)(-1)}=\frac{1}{(8)(18)-(20)(9)}$

$\frac{p}{-20+18}=\frac{q}{-9+8}=\frac{1}{146 -60}$

$\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}$

$p=\frac{1}{18},\:and\:q=\frac{1}{36}$

So,

$x=18\:and\:y=36.$

Q2 Formulate the following problems as a pair of equations and hence find their solutions: (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer:

Let the speed of the train and bus be u and v respectively

Now According to the question,

$\frac{60}{u}+\frac{240}{v}=4$

And

$\frac{100}{u}+\frac{200}{v}=4+\frac{1}{6}$

$\Rightarrow \frac{100}{u}+\frac{200}{v}=\frac{25}{6}$

Let,

$\frac{1}{u}=p\:and\:\frac{1}{v}=q$

Now, our equation becomes

$60p+140q=4$

$\Rightarrow 15p+60q=1.........(1)$

And

$100p+200q=\frac{25}{6}$

$\Rightarrow 4p+8q=\frac{1}{6}$

$\Rightarrow 24p+48q=1..........(2)$

By Cross Multiplication method,

$\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{q}{(60)(-1)-(48)(-1)}=\frac{p}{(-1)(24)-(-1)(15)}=\frac{1}{(15)(48)-(60)(24)}$

$\frac{p}{-60+48}=\frac{q}{-24+15}=\frac{1}{720-1440}$

$\frac{p}{-12}=\frac{q}{-9}=\frac{1}{-720}$

$p=\frac{12}{720}=\frac{1}{60},\:and\:q=\frac{9}{720}=\frac{1}{80}$

And Hence,

$x=60\:and\:y=80$

Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.7

Q1 The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Answer:

Let the age of Ani be $a$ , age of Biju be $b$ ,

Case 1: when Ani is older than Biju

age of Ani's father Dharam:

$d=2a$ and

age of his sister Cathy :

$c=\frac{b}{2}$

Now According to the question,

$a-b=3...........(1)$

Also,

$\\d-c=30 \\\Rightarrow 2a-\frac{b}{2}=30$

$\Rightarrow 4a-b=60..............(2)$

Now subtracting (1) from (2), we get,

$3a=60-3$

$\Rightarrow a=19$

putting this in (1)

$19-b=3$

$\Rightarrow b=16$

Hence the age of Ani and Biju is 19 years and 16 years respectively.

Case 2:

$b-a=3..........(3)$

And

$\\d-c=30 \\\Rightarrow 2a-\frac{b}{2}=30$

$\Rightarrow 4a-b=60..............(4)$

Now Adding (3) and (4), we get,

$3a=63$

$\Rightarrow a=21$

putting it in (3)

$b-21=3$

$\Rightarrow b=24.$

Hence the age of Ani and Biju is 21 years and 24 years respectively.

Q2 One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

[Hint : $x + 100 = 2(y - 100), y + 10 = 6(x - 10)$ ]

Answer:

Let the amount of money the first person and the second person having is x and y respectively

Noe, According to the question.

$x + 100 = 2(y - 100)$

$\Rightarrow x - 2y =-300...........(1)$

Also

$y + 10 = 6(x - 10)$

$\Rightarrow y - 6x =-70..........(2)$

Multiplying (2) by 2 we get,

$2y - 12x =-140..........(3)$

Now adding (1) and (3), we get

$-11x=-140-300$

$\Rightarrow 11x=440$

$\Rightarrow x=40$

Putting this value in (1)

$40-2y=-300$

$\Rightarrow 2y=340$

$\Rightarrow y=170$

Thus two friends had 40 Rs and 170 Rs respectively.

Q3 A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Now As we Know,

$speed=\frac{distance }{time}$

$\Rightarrow v=\frac{d}{t}$

$\Rightarrow d=vt..........(1)$
Now, According to the question,
$(v+10)=\frac{d}{t-2}$

$\Rightarrow (v+10){t-2}=d$

$\Rightarrow vt +10t-2v-20=d$

Now, Using equation (1), we have

$\Rightarrow -2v+10t=20............(2)$
Also,

$(v-10)=\frac{d}{t+3}$

$\Rightarrow (v-10)({t+3})=d$

$\Rightarrow vt+3v-10t-30=d$
$\Rightarrow3v-10t=30..........(3)$

Adding equations (2) and (3), we obtain:
$v=50.$
Substituting the value of x in equation (2), we obtain:
$(-2)(50)+10t=20$

$\Rightarrow -100+10t=20$

$\Rightarrow 10t=120$

$\Rightarrow t=12$
Putting this value in (1) we get,

$d=vt=(50)(12)=600$

Hence the distance covered by train is 600km.

Q4 The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:

Let the number of rows is x and the number of students in a row is y.
Total number of students in the class = Number of rows * Number of students in a row
$=xy$

Now, According to the question,

$\\xy = (x - 1) (y + 3) \\\Rightarrow xy= xy - y + 3x - 3 \\\Rightarrow 3x - y - 3 = 0 \\ \Rightarrow 3x - y = 3 ...... ... (1)$

Also,

$\\xy=(x+2)(y-3)\\\Rightarrow xy = xy + 2y - 3x - 6 \\ \Rightarrow 3x - 2y = -6 ... (2)$
Subtracting equation (2) from (1), we get:
$y=9$
Substituting the value of y in equation (1), we obtain:
$\\3x - 9 = 3 \\\Rightarrow 3x = 9 + 3 = 12 \\\Rightarrow x = 4$
Hence,
The number of rows is 4 and the Number of students in a row is 9.

Total number of students in a class

: $xy=(4)(9)=36$

Hence there are 36 students in the class.

Q5 In a $\Delta\textup{ABC}$ , $\angle C = 3 \angle B = 2( \angle A + \angle B)$ . Find the three angles.

Answer:

Given,

$\angle C = 3 \angle B = 2(\anlge A + \angle B)$

$\Rightarrow 3 \angle B = 2(\anlge A + \angle B)$

$\Rightarrow \angle B = 2 \angle A$

$\Rightarrow 2 \angle A -\angle B = 0..........(1)$

Also, As we know that the sum of angles of a triangle is 180, so

$\angle A +\angle B+ \angle C=180$

$\angle A +\angle B+ 3\angle B=180^0$

$\angle A + 4\angle B=180^0..........(2)$

Now From (1) we have

$\angle B = 2 \angle A.......(3)$

Putting this value in (2) we have

$\angle A + 4(2\angle A)=180^0.$

$\Rightarrow 9\angle A=180^0.$

$\Rightarrow \angle A=20^0.$

Putting this in (3)

$\angle B = 2 (20)=40^0$

And

$\angle C = 3 \angle B =3(40)=120^0$

Hence three angles of triangles $20^0,40^0\:and\:120^0.$

Q6 Draw the graphs of the equations $5x - y =5$ and $3x - 7 = 3$ . Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Answer:

Given two equations,

$5x - y =5.........(1)$

And

$3x - y = 3........(2)$

Points(x,y) which satisfies equation (1) are:

X	0	1	5
Y	-5	0	20

Points(x,y) which satisfies equation (1) are:

X	0	1	2
Y	-3	0	3

GRAPH:

Graph

As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).

Q7 Solve the following pair of linear equations: (i) $\\px + qy = p - q\\ qx - py = p + q$

Answer:

Given Equations,

$\\px + qy = p - q.........(1)\\ qx - py = p + q.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(q)(-p-q)-(q-p)(-p)}=\frac{y}{(q-p)(q)-(p)(-p-q)}=\frac{1}{(p)(-p)-(q)(q)}$

$\frac{x}{-p^2-q^2}=\frac{y}{p^2+q^2}=\frac{1}{-p^2-q^2}$

$x=1,\:and\:y=-1$

Q7 Solve the following pair of linear equations: (ii) $\\ax + by = c\\ bx +ay = 1 + c$

Answer:

Given two equations,

$\\ax + by = c.........(1)\\ bx +ay = 1 + c.........(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(b)(-1-c)-(-c)(a)}=\frac{y}{(-c)(b)-(a)(-1-c)}=\frac{1}{(a)(a)-(b)(b)}$

$\frac{x}{-b-bc+ac}=\frac{y}{-cb+a+ac}=\frac{1}{a^2-b^2}$

$x=\frac{-b-bc+ac}{a^2-b^2},\:and\:y=\frac{a-bc+ac}{a^2-b^2}$

Q7 Solve the following pair of linear equations: (iii) $\\\frac{x}{a} - \frac{y}{b} = 0\\ ax + by = a^2 +b^2$

Answer:

Given equation,

$\\\frac{x}{a} - \frac{y}{b} = 0\\\Rightarrow bx-ay=0...........(1)\\ ax + by = a^2 +b^2...............(2)$

Now By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-a)(-a^2-b^2)-(b)(0)}=\frac{y}{(0)(a)-(b)(-a^2-b^2)}=\frac{1}{(b)(b)-(-a)(a)}$

$\frac{x}{a(a^2+b^2)}=\frac{y}{b(a^2+b^2)}=\frac{1}{a^2+b^2}$

$x=a,\:and\:y=b$

Q7 Solve the following pair of linear equations: (iv) $\\(a-b)x + (a+b)y = a^2 -2ab - b^2\\ (a+b)(x+y) = a^2 +b^2$

Answer:

Given,

$\\(a-b)x + (a+b)y = a^2 -2ab - b^2..........(1)$

And

$\\ (a+b)(x+y) = a^2 +b^2\\\Rightarrow (a+b)x+(a+b)y=a^2+b^2...........(2)$

Now, Subtracting (1) from (2), we get

$(a+b)x-(a-b)x=a^2+b^2-a^2+2ab+b^2$

$\Rightarrow(a+b-a+b)x=2b^2+2ab$

$\Rightarrow 2bx=2b(b+2a)$

$\Rightarrow x=(a+b)$

Substituting this in (1), we get,

$(a-b)(a+b)+(a+b)y=a^2-2ab-b^2$

$\Rightarrow a^2-b^2+(a+b)y=a^2-2ab-b^2$

$\Rightarrow (a+b)y=-2ab$

$\Rightarrow y=\frac{-2ab}{a+b}$ .

Hence,

$x=(a+b),\:and\:y=\frac{-2ab}{a+b}$

Q7 Solve the following pair of linear equations: (v) $\\152x - 378y = -74\\ -378x + 152y = -604$

Answer:

Given Equations,

$\\152x - 378y = -74............(1)\\ -378x + 152y = -604............(2)$

As we can see by adding and subtracting both equations we can make our equations simple to solve.

So,

Adding (1) and )2) we get,

$-226x-226y=-678$

$\Rightarrow x+y=3...........(3)$

Subtracting (2) from (1) we get,

$530x-530y=530$

$\Rightarrow x-y=1...........(4)$

Now, Adding (3) and (4) we get,

$2x=4$

$\Rightarrow x=2$

Putting this value in (3)

$2+y=3$

$\Rightarrow y=1$

Hence,

$x=2\:and\:y=1$

Q8 ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

Answer:

As we know that in a quadrilateral the sum of opposite angles is 180 degrees.

So, From Here,

$4y+20-4x=180$

$\Rightarrow 4y-4x=160$

$\Rightarrow y-x=40............(1)$

Also,

$3y-5-7x+5=180$

$\Rightarrow 3y-7x=180........(2)$

Multiplying (1) by 3 we get,

$\Rightarrow 3y-3x=120........(3)$

Now,

Subtracting, (2) from (3) we get,

$4x=-60$

$\Rightarrow x=-15$

Substituting this value in (1) we get,

$y-(-15)=40$

$\Rightarrow y=40-15$

$\Rightarrow y=25$

Hence four angles of a quadrilateral are :

$\angle A =4y+20=4(25)+20=100+20=120^0$

$\angle B =3y-5=3(25)-5=75-5=70^0$

$\angle C =-4x=-4(-15)=60^0$

$\angle D =-7x+5=-7(-15)+5=105+5=110^0$

NCERT solutions for class 10 maths chapter wise

Chapter No.	Chapter Name
Chapter 1	NCERT solutions for class 10 maths chapter 1 Real Numbers
Chapter 2	NCERT solutions for class 10 maths chapter 2 Polynomials
Chapter 3	NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables
Chapter 4	NCERT solutions for class 10 maths chapter 4 Quadratic Equations
Chapter 5	NCERT solutions for class 10 chapter 5 Arithmetic Progressions
Chapter 6	NCERT solutions for class 10 maths chapter 6 Triangles
Chapter 7	NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
Chapter 8	NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
Chapter 9	NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
Chapter 10	NCERT solutions class 10 maths chapter 10 Circles
Chapter 11	NCERT solutions for class 10 maths chapter 11 Constructions
Chapter 12	NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles
Chapter 13	NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes
Chapter 14	NCERT solutions for class 10 maths chapter 14 Statistics
Chapter 15	NCERT solutions for class 10 maths chapter 15 Probability

NCERT solutions of class 10 subject wise

NCERT solutions for class 10 maths

NCERT solutions for class 10 science

How to use NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables ?

NCERT solutions are the most important tool when you are appearing for board examinations. 90% paper of CBSE board examinations, directly come from the NCERT.
Now you have done the NCERT solutions for class 10 maths chapter 2 polynomials and learned the approach to solving questions in the step by step method.
After covering NCERTs you should target the past year papers of CBSE board examinations. The previous year papers will cover the rest 10% part of the class 10 board exams.

Keep working hard & happy learning!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Question: What is the weightage of the chapter pair of linear equations for CBSE board exam ?

Answer:

CBSE doesn't provides the marks distributions chapter-wise but it provides the total weightage of a unit( upto 4-5 chapters). As per CBSE the total weightage of algebra ( 4 chapters) is 20 marks in the final board exam.

Question: Where can I find the complete solutions of NCERT class 10 maths ?

Answer:

Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.

Question: What are the important topics of class 10 maths chapter 3 pairs of linear equations in two variables ?

Answer:

Formation of linear equations using statements, algebraic interpretation of linear equations, solving a linear equation with two variables, representation of linear equation in a graph, and solutions of linear equations using the graph are important topics from this chapter.

Question: How many chapters are there in the class 10 maths ?

Answer:

There are 15 chapters in the class 10 maths NCERT. Chapter 1- Real Numbers, Chapter 2- Polynomials, Chapter 3- Pair of Linear Equations in Two Variables, Chapter 4- Quadratic Equations, Chapter 5- Arithmetic Progressions, Chapter 6- Triangles, Chapter 7- Coordinate Geometry, Chapter 8- Introduction to Trigonometry, Chapter 9- Some Applications of Trigonometry, Chapter 10- Circles, Chapter 11- Constructions, Chapter 12- Areas Related to Circles, Chapter 13- Surface Areas and Volumes, Chapter 14- Statistics, Chapter 15- Probability are the chapters in the NCERT class 10 maths.

Question: Which is the official website of NCERT ?

Answer:

http://ncert.nic.in/ is the official website of the NCERT where you can get NCERT textbooks from class 1 to 12 and syllabus from class 1 to 12 for all the subjects.

Question: How the NCERT solutions are helpful in the board exam ?

Answer:

NCERT solutions are not only important when you stuck while solving the problems but you will get new ways to solve the problems. As these solutions are provided in a very simple language, so it can be understood by an average student also. Students are advised to go through these solutions, so you will get how to answer in the board exam in order to get good marks in the board exam.

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.1

Answer:

NCERT solutions for class 10 maths Chapter 3 Pair of Linear Equations in two variables Excercise: 3.2

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Given the equation,

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NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.3

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.4

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.5

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NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in two variables Excercise: 3.6

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