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

Class 10 Maths Chapter 3 solutions 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,

$\begin{array}{rl}& (x-7)=7\times (y-7)\\ & \Rightarrow x-7=7y-49\\ & \Rightarrow x-7y=-42\dots .1\end{array}$

Also,

$\begin{array}{rl}& (x+3)=3(y+3)\\ & \Rightarrow x+3=3y+9\\ & \Rightarrow x-3y=6\dots .(2)\end{array}$

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

And From (2) we get,

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

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

GRAPH:

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,

$\begin{array}{rl}& 3x+6y=3900\dots .(1)\\ & x+3y=1300\dots .(2)\end{array}$

From(1) we have

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

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

Now, From (2), we have,

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

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

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:

$2x+y=160\dots .$

After One Month:

$4x+2y=300$

Now, From (1) we have

$y=160-2x$

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

And From (2) we have,

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

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

Graph:

Class 10 Maths Chapter 3 solutions 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.

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

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

$\begin{array}{rl}& x=y+4\\ & \Rightarrow x-y=4.\end{array}$

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

Different points (x,y) satisfying (2)

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

$5x+7y=50\dots \dots$

And

$7x+5y=46\dots \dots$

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

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

The 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 ${\sqrt[|\frac{a_1}{a_2}]{|\frac{b_1}{b_2}}]}_{\text{and }}\left|\frac{c_1}{c_2}\right|$find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) $\begin{array}{rl}& 5x-4y+8=0\\ & 7x+6y-9=0\end{array}$

Answer:

Give, Equations,

$\begin{array}{rl}& 5x-4y+8=0\\ & 7x+6y-9=0\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

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

As we can see

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

It means that both lines intersect at exactly one point.

Q2 On comparing the ratios $\left[\frac{a_1}{a_2}\right],\frac{b_1}{b_2}$ 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: (ii) $\begin{array}{rl}& 9x+3y+12=0\\ & 18x+6y+24=0\end{array}$

Answer:

Given, Equations,

$\begin{array}{rl}& 9x+3y+12=0\\ & 18x+6y+24=0\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\text{ and }\\ & \frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\end{array}$

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.

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: (iii) $\begin{array}{rl}& 6x-3y+10=0\\ & 2x-y+9=0\end{array}$

Answer:

Give, Equations,

$\begin{array}{rl}& 6x-3y+10=0\\ & 2x-y+9=0\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

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

As we can see

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

It means that both lines are parallel to each other.

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

Answer:

Give, Equations,

$\begin{array}{rl}& 3x+2y=5\\ & 2x-3y=7\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

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

As we can see

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

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

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

Answer:

Given, Equations,

$\begin{array}{rl}& 2x-3y=8\\ & 4x-6y=9\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\text{ and }\\ & \frac{c_1}{c_2}=\frac{8}{9}\end{array}$

As we can see

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

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

Q3 On comparing the ratios $\left[\frac{a_1}{a_2}\right],\left[\frac{b_1}{b_2}\right]$ and $\left[\frac{c_1}{c_2}\right]$, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent: (iii) $\frac{3}{2}x+\frac{5}{3}y=7;9x-10y=14$

Answer:

Given, Equations,

$\begin{array}{rl}& \frac{3}{2}x+\frac{5}{3}y=7\\ & 9x-10y=14\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{18}=\frac{1}{6}\\ & \frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-30}=-\frac{1}{6}\text{ and }\\ & \frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}\end{array}$

As we can see

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

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: (iv) $5x-3y=11;-10x+6y=-22$

Answer:

Given, Equations,

$\begin{array}{rl}5x-3y& =11\\ -10x+6y& =-22\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\text{ and }\\ & \frac{c_1}{c_2}=\frac{11}{-22}=-\frac{1}{2}\end{array}$

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 $\frac{a_1}{a_2}|,[\frac{b_1}{\frac{b_2}{2}}|\underline{\begin{array}{l}\frac{c_1}{c_2}\end{array}}$ find out whether the following pair of linear equations are consistent, or inconsistent (v) $\frac{4}{3}x+2y=8;2x+3y=12$

Answer:

Given, Equations,

$\begin{array}{rl}& \frac{4}{3}x+2y=8\\ & 2x+3y=12\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{4/3}{2}=\frac{4}{6}\\ & \frac{b_1}{b_2}=\frac{2}{3}\text{ and }\\ & \frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}\end{array}$

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) $x+y=52x+2y=10$

Answer:

Given, Equations,

$\begin{array}{rl}& x+y=5\\ & 2x+2y=10\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{1}{2}\text{ and }\\ & \frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}\end{array}$

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

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

Answer:

Given, Equations,

$\begin{array}{rl}& x-y=8\\ & 3x-3y=16\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{1}{3}\\ & \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\text{ and }\\ & \frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}\end{array}$

As we can see

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

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) $2x+y-6=0,4x-2y-4=0$

Answer:

Given, Equations,

$\begin{array}{rl}& 2x+y-6=0\\ & 4x-2y-4=0\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\text{ and }\\ & \frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\end{array}$

As we can see

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

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,

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

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) $2x-2y-2=0,4x-4y-5=0$

Answer:

Given, Equations,

$\begin{array}{rl}& 2x-2y-2=0\\ & 4x-4y-5=0\end{array}$

Comparing these equations with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

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

$\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\text{ and }$

$\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$

As we can see

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

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.

$\begin{array}{rl}& l=b+4\\ & l-b=4\end{array}$

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

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

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

$\begin{array}{rl}& l+b+l-b=4+36\\ & \Rightarrow 2l=40\\ & \Rightarrow l=20\end{array}$

Putting this in equation (1),

$\begin{array}{rl}& \Rightarrow 20-b=4\\ & \Rightarrow b=20-4\\ & \Rightarrow b=16\end{array}$

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 $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, is,

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

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

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{3}{3}=1\end{array}$

As

$\frac{1}{2}\ne 1$

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 $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, for being parallel is,

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

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

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\\ & \frac{c_1}{c_2}=\frac{-8}{-8}=1\end{array}$

As
$\frac{1}{2}=\frac{1}{2}\ne 1$ 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 $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, 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,

$\begin{array}{rl}& \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\ & \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\\ & \frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\end{array}$

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

Q7 Draw the graphs of the equations $x-y+1=0$and $3x+2y-12=0$ . 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,

$x-y+1=0$

And

$3x+2y-12=0$

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

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

GRAPH:

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.

Pair of linear equations in two variables class 10 solutions Excercise: 3.3

Q1 Solve the following pair of linear equations by the substitution method. (i) $\begin{array}{rl}& x+y=14\\ & x-y=4\end{array}$

Answer:

Given, two equations,

$\begin{array}{rl}& x+y=14\dots \dots .(1)\\ & x-y=4\dots \dots .(2)\end{array}$

Now, from (1), we have

$y=14-x$

Substituting this in (2), we get

$\begin{array}{rl}& x-(14-x)=4\\ & \Rightarrow x-14+x=4\\ & \Rightarrow 2x=4+14=18\\ & \Rightarrow x=9\end{array}$

Substituting this value of $x$ in (3)

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

Hence, Solution of the given equations is $\mathrm{x}=9$ and $\mathrm{y}=5$.

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

Answer:

Given, two equations,

$\begin{array}{rl}& s-t=3\dots \\ & \frac{s}{3}+\frac{t}{2}=6.\end{array}$

Now, from (1), we have

$s=t+3.$

Substituting this in (2), we get

$\begin{array}{rl}& \frac{t+3}{3}+\frac{t}{2}=6\\ & \Rightarrow \frac{2t+6+3t}{6}=6\\ & \Rightarrow 5t+6=36\\ & \Rightarrow 5t=30\\ & \Rightarrow t=6\end{array}$

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) $\begin{array}{rl}& 3x-y=3\\ & 9x-3y=9\end{array}$

Answer:

Given, two equations,

$\begin{array}{rl}& 3x-y=3\dots \dots \\ & 9x-3y=9\dots .\end{array}$

Now, from (1), we have

$y=3x-3..$

Substituting this in (2), we get

$\begin{array}{rl}& 9x-3(3x-3)=9\\ & \Rightarrow 9x-9x+9=9\\ & \Rightarrow 9=9\end{array}$

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) $\begin{array}{rl}& 0.2x+0.3y=1.3\\ & 0.4x+0.5y=2.3\end{array}$

Answer:

Given, two equations,

$\begin{array}{rl}& 0.2x+0.3y=1.3\\ & 0.4x+0.5y=2.3\end{array}$

Now, from (1), we have

$y=\frac{1.3-0.2x}{0.3}$

Substituting this in (2), we get

$\begin{array}{rl}& 0.4x+0.5\left(\frac{1.3-0.2x}{0.3}\right)=2.3\\ & \Rightarrow 0.12x+0.65-0.1x=0.69\\ & \Rightarrow 0.02x=0.69-0.65=0.04\\ & \Rightarrow x=2\end{array}$

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,

$x=2\text{ and }y=3$

Q1 Solve the following pair of linear equations by the substitution method. (v) \begin{tabular}{|l|}$\begin{array}{rl}& \sqrt{2}x+\sqrt{3}y=0\\ & \sqrt{3}x-\sqrt{8}y=0\end{array}$

Answer:

Given, two equations,

$\begin{array}{rl}& \sqrt{2}x+\sqrt{3}y=0\\ & \sqrt{3}x-\sqrt{8}y=0\end{array}$

Now, from (1), we have

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

Substituting this in (2), we get

$\begin{array}{rl}& \sqrt{3}x-\sqrt{8}(-\frac{\sqrt{2}x}{\sqrt{3}})=0\\ & \Rightarrow \sqrt{3}x=\sqrt{8}(-\frac{\sqrt{2}x}{\sqrt{3}})\\ & \Rightarrow 3x=-4x\\ & \Rightarrow 7x=0\\ & \Rightarrow x=0\end{array}$

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 , y = 0 .

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

Answer:

Given,

$\begin{array}{rl}& \frac{3x}{2}-\frac{5y}{3}=-2.\\ & \frac{x}{3}+\frac{y}{2}=\frac{13}{6}\dots \end{array}$

From (1) we have,

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

Putting this in (2) we get,

$\begin{array}{rl}& \frac{1}{3}\times \frac{2}{3}(\frac{5y}{3}-2)+\frac{y}{2}=\frac{13}{6}\\ & \frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}\\ & \frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}\\ & \frac{47y}{54}=\frac{13}{6}+\frac{4}{9}\\ & \frac{47y}{54}=\frac{117}{54}+\frac{24}{54}\\ & 47y=117+24\\ & 47y=141\end{array}$

$\begin{array}{rl}& y=\frac{141}{47}\\ & y=3\end{array}$

putting this value in (3) we get,

$\begin{array}{rl}& x=\frac{2}{3}(\frac{5y}{3}-2)\\ & x=\frac{2}{3}(\frac{5\times 3}{3}-2)\\ & x=\frac{2}{3}(5-2)\\ & x=\frac{2}{3}\times 3\\ & x=2\end{array}$

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,

$\begin{array}{rl}& 2x+3y=11\dots \dots \\ & 2x-4y=-24\dots \end{array}$

Now, from (1), we have

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

Substituting this in (2), we get

$\begin{array}{rl}& 2x-4\left(\frac{11-2x}{3}\right)=-24\\ & \Rightarrow 6x-44+8x=-72\\ & \Rightarrow 14x=44-72\\ & \Rightarrow 14x=-28\\ & \Rightarrow x=-2\end{array}$

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,

$x=-2,\text{ and }y=5$

Now,

As it satisfies $y=mx+3$,

$\begin{array}{rl}& \Rightarrow 5=m(-2)+3\\ & \Rightarrow 2m=3-5\\ & \Rightarrow 2m=-2\\ & \Rightarrow m=-1\end{array}$

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,

$y-x=26.$

And

$y=3x.$

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

$\begin{array}{rl}& 3x-x=26\\ & \Rightarrow 2x=26\\ & \Rightarrow x=13\end{array}$

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,

$x+y=180$

Also given in the question,

$x-y=18$

Now, From (2) we have,

$y=x-18$

Substituting this value in (1)

$\begin{array}{rl}& x+x-18=180\\ & \Rightarrow 2x=180+18\\ & \Rightarrow 2x=198\\ & \Rightarrow x=99\end{array}$

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

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

Hence the two supplementary angles are

${99}^0\text{ and }{81}^0$

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,

$\begin{array}{rl}& 7x+6y=3800..\\ & 3x+5y=1750..\end{array}$

Now, From (1) we have

$y=\frac{3800-7x}{6}$

Substituting this value of $y$ in (2)

$\begin{array}{rl}& 3x+5\left(\frac{3800-7x}{6}\right)=1750\\ & \Rightarrow 18x+19000-35x=1750\times 6\\ & \Rightarrow -17x=10500-19000\\ & \Rightarrow -17x=-8500\\ & \Rightarrow x=\frac{8500}{17}\\ & \Rightarrow x=500\end{array}$

Now, Substituting this value of x in (3)

$y=\frac{3800-7x}{6}=\frac{3800-7\times 500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$

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

$y=\frac{3800-7x}{6}=\frac{3800-7\times 500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$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,

$\begin{array}{rl}& \frac{x+2}{y+2}=\frac{9}{11}\\ & \Rightarrow 11(x+2)=9(y+2)\\ & \Rightarrow 11x+22=9y+18\\ & \Rightarrow 11x-9y=-4\dots \dots \dots \end{array}$

Also,

$\begin{array}{rl}& \frac{x+3}{y+3}=\frac{5}{6}\\ & \Rightarrow 6(x+3)=5(y+3)\\ & \Rightarrow 6x+18=5y+15\\ & \Rightarrow 6x-5y=-3\dots \dots \dots \end{array}$

Now, From (1) we have

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

Substituting this value of y in (2)

$\begin{array}{rl}& 6x-5\left(\frac{11x+4}{9}\right)=-3\\ & \Rightarrow 54x-55x-20=-27\\ & \Rightarrow -x=20-27\\ & \Rightarrow x=7\end{array}$

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

$\begin{array}{rl}& x+5=3(y+5)\\ & \Rightarrow x+5=3y+15\\ & \Rightarrow x-3y=10\dots \dots \end{array}$

Also,

$\begin{array}{rl}& x-5=7(y-5)\\ & \Rightarrow x-5=7y-35\\ & \Rightarrow x-7y=-30\dots \end{array}$

Now,
From (1) we have,

$x=10+3y$

Substituting this value of $x$ in (2)

$\begin{array}{rl}& 10+3y-7y=-30\\ & \Rightarrow -4y=-30-10\end{array}$

$\begin{array}{rl}& \Rightarrow 4y=40\\ & \Rightarrow y=10\end{array}$

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.

Pair of linear equations in two variables class 10 solutions Excercise: 3.4

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

(i) $x+y=5$ and $2x-3y=4$

Answer:
Elimination Method:
Given, equations

$\begin{array}{rl}& x+y=5\dots \dots \dots ..(1)\text{ and }\\ & 2x-3y=4\dots \dots .(2)\end{array}$

Now, multiplying (1) by 3 we, get

$3x+3y=15$

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

$\begin{array}{rl}& 2x-3y+3x+3y=4+15\\ & \Rightarrow 5x=19\\ & \Rightarrow x=\frac{19}{5}\end{array}$

Substituting this value in (1) we, get

$\begin{array}{rl}& \frac{19}{5}+y=5\\ & \Rightarrow y=5-\frac{19}{5}\\ & \Rightarrow y=\frac{6}{5}\end{array}$

Hence,

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

Substitution method :
Given, equations

$\begin{array}{rl}& x+y=5\dots \\ & 2x-3y=4.\end{array}$

Now, from (1) we have,

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

substituting this value in (2)

$\begin{array}{rl}& 2x-3(5-x)=4\\ & \Rightarrow 2x-15+3x=4\\ & \Rightarrow 5x=19\\ & \Rightarrow x=\frac{19}{5}\end{array}$

Substituting this value of $x$ in (3)

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

Hence,

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

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

(ii) $3x+4y=10$ and $2x-2y=2$