NCERT Solutions for Exercise 3.7 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

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NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7

NCERT Solutions for class 10 maths ex 3.7 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.7 class 10 includes cross-multiplication, elimination, and substitution methods for solving two-variable linear equations, as well as all of the topics from the whole chapter. A linear equation in two variables is a system of equations of the form ax + by + c = 0, where a, b, c are real numbers, having unique solution, no solutions, or infinitely many solutions. To identify the set of solutions to the linear equations with two variables, there are five mathematical ways including graphical method, replacement method, cross Multiplication method, and elimination method.

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Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.7
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NCERT Solutions for Exercise 3.7 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

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Pair of Linear Equations in Two Variables Class 10 Chapter 3 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 (i) Solve the following pair of linear equations: $\\px + qy = p - q\\ qx - py = p + q$

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

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

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

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

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 :

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

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

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

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

Benefits of NCERT Solutions for Class 10 Maths Exercise 3.7:

NCERT Solutions for Class 10 Maths exercise 3.7 extensively covers a wide range of inquiries identified with every one of the points like cross-multiplication, elimination, and substitution methods.
Exercise 3.7 Class 10 Maths, helps the understudies who wish to score high in CBSE Term tests are encouraged to through NCERT solutions for Class 10 Maths exercise 3.7 all through.
By tackling the NCERT solution for Class 10 Maths chapter 3 exercise 3.7, students can accomplish a decent score in term tests just as in cutthroat assessments.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. According to NCERT solutions for Class 10 Maths chapter 3 exercise 3.7 , What strategies are employed to obtain the solution of a pair of linear equations?

Option 1:

• Graphical method

• Algebraic methods

Method of Substitution
Method of Elimination
Method of cross-multiplication

are the approaches used to find the solution to a pair of linear equations

2. If the pair of linear equations has an infinite number of solutions, then the equations will be _______

If the pair of linear equations have an infinite number of solutions, then the equations will be consistent and dependent.

3. The pair of equations x=m and y=n graphically represents lines which are

Parallel to each other
Intersecting at (n, m)
coincident each other
Intersecting at (m, n)

The pair of equations x=m and y=n graphically represents lines that are intersecting at (m, n)

Option (d) intersecting at (m, n)

4. What possibilities exist when two lines intersect in a plane, and what are they?

When two lines are in a plane, there are three alternative solutions. They really are.

• Two lines may intersect at times.

• Two lines may not intersect at times, and they may be parallel to each other.

• Two lines may be coincident at times.

5. Is it important to learn all three methods to solve pairs of linear equations in two variables of NCERT solutions for Class 10 Maths chapter 3 exercise 3.7 ?

Yes, it is important to learn all three methods to solve pairs of linear equations in two variables.

6. What types of questions are covered in the NCERT solutions for Class 10 Maths chapter 3 exercise 3.7 ?

NCERT solutions for Class 10 Maths chapter 3 exercise 3.7 Consists of questions covering topics from the overall chapter that is it has problems related to all the topics like cross-multiplication, elimination, and substitution methods.

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sai raja 01 September,2024

show the previous year exams ?

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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best of luck!

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i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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NCERT Solutions for Exercise 3.7 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.7

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Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.7

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