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

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

Edited By Ramraj Saini | Updated on Nov 13, 2023 03:31 PM IST | #CBSE Class 10th

## 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.

These class 10 maths ex 3.2 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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

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.

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

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.

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.

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.

Given,

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 $20^0,40^0\:and\:120^0.$

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:

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

Given Equations,

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

Now By Cross multiplication method,

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

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

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

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

Given two equations,

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

Now By Cross multiplication method,

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

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

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

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

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,

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

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

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

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

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,

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

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,

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

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 :

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

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

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

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

## More About NCERT Solutions for Class 10 Maths Exercise 3.7

NCERT solutions for Class 10 Maths exercise 3.7 covered the graphical approach and algebraic method of linear equations in two variables, types of solutions, and their graphs. At times, two systems of equations can be parallel. In such situations, they have no recourse. The two systems of equations may coincide in exceptional cases, resulting in an infinite number of solutions for the given system. The provided system is said to be consistent since it has a unique solution if the two systems of equations overlap. The following tasks are included along with NCERT book Class 10 Maths chapter 3 exercisr 3.7. All 8 questions are related to the graphical method and algebraic method are given in Exercise 3.7 Class 10 Maths. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.

## 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.

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## Subject Wise NCERT Exemplar Solutions

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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### Questions related to CBSE Class 10th

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Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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