# 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: It is an important chapter of algebra where you will learn to solve the linear equation having two variables. In solutions of NCERT 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. CBSE 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. The problems related to this chapter are very interesting and are important in the board exam and also for many other competitive exams. CBSE 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. Cross-multiplication method to solve the linear equations is also given in the NCERT Solutions of this chapter.

Seven exercises of this chapter are explained below.

Exercise:3.1

Exercise:3.2

Exercise:3.3

Exercise:3.4

Exercise:3.5

Exercise:3.6

Exercise:3.7

## 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

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:

## 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.

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.

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:

## 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.

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.

## (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.

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,

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

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.

## 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: (ii)    $\\9x + 3y + 12 = 0\\ 18x + 6y + 24 = 0$

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.

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.

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)    $2x - 3y = 8;\qquad 4x - 6y = 9$

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.

## 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: (iii)    $\frac{3}{2}x + \frac{5}{3}y = 7;\qquad 9x -10y = 14$

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.

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

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.

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

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.

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

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.

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.

### 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,

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

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

As

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

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,

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

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

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

As

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

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,

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

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

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

As

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

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.

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

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.

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)    $\\ 3 x - y = 3\\ 9x - 3y = 9$

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

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

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

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 $x=2\:\:and\:\:y=3.$

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.

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

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.

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

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

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.

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.

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

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

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

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

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,

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

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

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,

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

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,

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

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,

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

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

$\frac{3}{5}$

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.

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.

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.

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

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

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)    $\\2x + y = 5 \\ 3x + 2y = 8$

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) $\\3x -5 y = 20\\ 6x - 10y = 40$

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

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,

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

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

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

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

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

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, $a=5\:and\:b=1$.

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.

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

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

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

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

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

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

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.

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

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

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 $x+y=15+5=20.$

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.

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.

## Pair of Linear Equations in two variables Excercise: 3.6

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

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,

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

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

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

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

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

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,

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

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

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,

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

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

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,

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

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

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

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,

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

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

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,

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

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

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,

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

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

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,

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

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

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

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

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 $20^0,40^0\:and\:120^0.$

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).

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

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 :

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

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

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

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

## NCERT solutions for class 10 maths chapter wise

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

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

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