NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables 2025-26

Exercise 3.1: This exercise involves solving word problems by setting up a pair of linear equations. Students will need to translate word problems into mathematical equations and solve them to find the solution.

Exercise 3.2: This exercise covers the concept of cross-multiplication and its application in solving linear equations. Students will learn to solve equations using the cross-multiplication method and will also solve problems related to this topic.

Exercise 3.3: This exercise involves solving equations that are reducible to a pair of linear equations in two variables. Students will learn to reduce equations to linear form and solve them.

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

Exercise 3.1

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

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

Ans (i): Assuming that the number of girls and boys be $$x$$ and $$y$$ respectively.

Writing the algebraic representation using the information given in the question-

$$x+y=10$$

$$x-y=4$$

Solution table for $$x+y=10$$-

$$x=10-y$$

Solution table for $$x-y=4$$-

$$x=4+y$$

Graphical representation-

As we can see from the graph above, the point of intersection for the lines is $$(7,3)$$. Therefore, we can say that there are $$7$$ girls and $$3$$ boys in the class.

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

Ans (ii): Assuming that the cost of $$1$$ pencil and $$1$$ pen be $$x$$ and $$y$$ respectively.

Writing the algebraic representation using the information given in the question-

$$5x+7y=50$$

$$7x+5y=46$$

Solution table for $$5x+7y=50$$-

$$x=\frac{50-7y}{5}$$

Solution table for $$7x+5y=46$$-

$$x=\frac{46-5y}{7}$$

Graphical representation-

As we can see from the graph above, the point of intersection for the lines is $$(3,5)$$. Therefore, we can say that the cost of a pencil is $$Rs3$$ and the cost of a pen is $$Rs\text{ 5}$$.

2. 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 at a point, are parallel or coincident-

(i) $$\mathbf{5}\mathbf{x}-\mathbf{4}\mathbf{y}+\mathbf{8}=\mathbf{0}$$

$$\mathbf{7}\mathbf{x}+\mathbf{6}\mathbf{y}-\mathbf{9}=\mathbf{0}$$

Ans: $$5x-4y+8=0$$

$$7x+6y-9=0$$

Calculating the values of $$a_1,b_1,c_1,a_2,b_2$$ and $$c_2$$ by comparing the above equations with $$a_{1}x+b_{1}y+c_1=0$$ and $$a_{2}x+b_{2}y+c_2=0$$,

$$a_1=5$$, $$b_1=-4$$, $$c_1=8$$

$$a_2=7$$, $$b_2=6$$, $$c_2=-9$$

$$\frac{a_1}{a_2}=\frac{5}{7}$$

$$\frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}$$

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

Therefore, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point.

(ii) $$\mathbf{9}\mathbf{x}+\mathbf{3}\mathbf{y}+\mathbf{12}=\mathbf{0}$$

$$\mathbf{18}\mathbf{x}+\mathbf{6}\mathbf{y}+\mathbf{24}=\mathbf{0}$$

Ans (ii): Calculating the values of $$a_1,b_1,c_1,a_2,b_2$$ and $$c_2$$ by comparing the above equations with $$a_{1}x+b_{1}y+c_1=0$$ and $$a_{2}x+b_{2}y+c_2=0$$,

$$a_1=9$$, $$b_1=3$$, $$c_1=12$$

$$a_2=18$$, $$b_2=6$$, $$c_2=24$$

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

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

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

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

Therefore, the lines representing the given pair of equations have infinite solutions as they are coincident.

(iii) $$\mathbf{6}\mathbf{x}-\mathbf{3}\mathbf{y}+\mathbf{10}=\mathbf{0}$$

$$\mathbf{2}\mathbf{x}-\mathbf{y}+\mathbf{9}=\mathbf{0}$$

Ans (iii): Calculating the values of $$a_1,b_1,c_1,a_2,b_2$$ and $$c_2$$ by comparing the above equations with $$a_{1}x+b_{1}y+c_1=0$$ and $$a_{2}x+b_{2}y+c_2=0$$,

$$a_1=6$$, $$b_1=-3$$, $$c_1=10$$

$$a_2=2$$, $$b_2=-1$$, $$c_2=9$$

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

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

$$\frac{c_1}{c_2}=\frac{10}{9}$$

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

Therefore, the lines representing the given pair of equations have no solutions as they are parallel to each other.

3. 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 pairs of linear equations are consistent, or inconsistent.

(i) $$3\mathbf{x}+2\mathbf{y}=5;2x-3y=7$$

Ans (i): For the given equations-

 $$\frac{a_1}{a_2}=\frac{3}{2}$$

$$\frac{b_1}{b_2}=\frac{-2}{3}$$

$$\frac{c_1}{c_2}=\frac{5}{7}$$

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

So, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point. 

Therefore, the given pair of lines is consistent.

(ii) $$2x-3y=8;4\mathbf{x}-6\mathbf{y}=9$$

Ans (ii): For the given equations-

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

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

$$\frac{c_1}{c_2}=\frac{8}{9}$$

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

So, the lines representing the given pair of equations have no solutions as they are parallel to each other.

Therefore, the given pair of lines is inconsistent.

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

Ans (iii): For the given equations-

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

$$\frac{b_1}{b_2}=\frac{-1}{6}$$

$$\frac{c_1}{c_2}=\frac{1}{4}$$

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

So, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point. 

Therefore, the given pair of lines is consistent.

(iv) $$5x-3y=11;-10\mathbf{x}+6\mathbf{y}=-22$$

Ans (iv): For the given equations-

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

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

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

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

So, the lines representing the given pair of equations have infinite number of solutions as they are coincident.

Therefore, the given pair of lines is consistent.

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

Ans (v): For the given equations-

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

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

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

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

So, the lines representing the given pair of equations have an infinite number of solutions as they are coincident.

Therefore, the given pair of lines is consistent.

4. Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically-

(i) $$x+y=5;2\mathbf{x}+2\mathbf{y}=10$$

Ans (i): For the given equations-

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

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

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

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

So, the lines representing the given pair of equations have infinite number of solutions as they are coincident.

Therefore, the given pair of lines is consistent.

Solution table for $$x+y=5$$-

$$x=5-y$$

Solution table for $$2x+2y=10$$-

$$x=\frac{10-2y}{2}$$

Graphical representation-

As shown in the graph above, the two lines are overlapping each other. Hence, they have infinite solutions.

(ii) $$x-y=8;3\mathbf{x}-3\mathbf{y}=16$$

Ans (ii): For the given equations-

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

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

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

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

So, the lines representing the given pair of equations have no solutions as they are parallel to each other.

Therefore, the given pair of lines is inconsistent.

(iii) $$2x+y-6=0;4\mathbf{x}-2\mathbf{y}-4=0$$

Ans (iii): For the given equations-

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

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

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

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

So, the given pair of equations have a unique solution, that is, the lines intersect at exactly one point. 

Therefore, the given pair of lines is consistent.

Solution table for $$2x+y-6=0$$-

$$y=6-2x$$

Solution table for $$4x-2y-4=0$$-

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

Graphical representation-

As shown in the graph above, the two lines intersect each other at only one point $$(2,2)$$.

(iv) $$2x-2y-2=0;4\mathbf{x}-4\mathbf{y}-5=0$$

Ans (iv):  For the given equations-

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

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

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

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

So, the lines representing the given pair of equations have no solutions as they are parallel to each other.

Therefore, the given pair of lines is inconsistent.

5. Half the perimeter of a rectangular garden, whose length is $$\mathbf{4}\mathbf{m}$$ more than its width, is $$\mathbf{36}\mathbf{m}$$. Find the dimensions of the garden.

Ans: Assuming that the width and length of the garden be $$x$$ and $$y$$ respectively.

Writing the algebraic representation using the information given in the question-

$$y-x=4$$

$$x+y=36$$

Solution table for $$y-x=4$$-

$$y=x+4$$

Solution table for $$x+y=36$$-

$$y=36-x$$

Graphical representation-

As shown in the graph above, the two lines intersect each other at only one point $$(16,20)$$. Therefore the length of the garden is $$20m$$ and its breadth is $$\text{16 }m$$.

6. Given the linear equation $$\mathbf{2}\mathbf{x}+\mathbf{3}\mathbf{y}-\mathbf{8}=\mathbf{0}$$, write another linear equations in two variables such that the geometrical representation of the pair so formed is-

(i) Intersecting lines

Ans (i): If two lines are intersecting then-

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

So, taking the second line as $$2x+4y-6=0$$,

Now, 

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

$$\frac{b_1}{b_2}=\frac{3}{4}$$

As $$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$$, the two lines are interesting each other.

(ii) Parallel lines

Ans: If two lines are intersecting then-

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

So, taking the second line as $$4x+6y-8=0$$,

Now, 

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

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

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

As $$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$$, the two lines are parallel to each other.

(iii) Coincident lines

Ans: If two lines are intersecting then-

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

So, taking the second line as $$6x+9y-24=0$$,

Now, 

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

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

$$\frac{c_1}{c_2}=\frac{1}{3}$$

As $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$, the two lines are coincident.

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

Ans: Solution table for $$x-y+1=0$$-

$$x=y-1$$

Solution table for $$3x+2y-12=0$$-

$$x=\frac{12-2y}{3}$$

Graphical representation-

As shown in the graph above, the lines are intersecting each other at point $$(2,3)$$ and $$x-axis$$ at $$(-1,0)$$ and $$(4,0)$$. So the obtained triangle has vertices $$(2,3)$$, $$(-1,0)$$ and $$(4,0)$$.

Exercise 3.2

1. Solve the following pair of linear equations by the substitution method.

(i) $$x+y=14;x-y=4$$

Ans (i): The given equations are-

$$x+y=14$$           …… (i)

$$x-y=4$$             …… (ii)

From equation (i)-

$$x=14-y$$          …… (iii)

Substituting (iii) in equation (ii), we get

$$(14-y)-y=4$$

$$14-2y=4$$

$$10=2y$$

$$y=5$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=9$$

Therefore, $$x=9$$ and $$y=5$$.

(ii) $$s-t=3;\frac{s}{3}+\frac{t}{2}=6$$

Ans (ii): The given equations are-

$$s-t=3$$           …… (i)

$$\frac{s}{3}+\frac{t}{2}=6$$       …… (ii)

From equation (i)-

$$s=t+3$$          …… (iii)

Substituting (iii) in equation (ii), we get

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

$$2t+6+3t=36$$

$$5t=30$$

$$t=6$$                  …… (iv)

Substituting (iv) in (iii), we get

$$s=9$$

Therefore, $$s=9$$ and $$t=6$$.

(iii) $$3x-y=3;9x-3y=9$$

Ans (iii): The given equations are-

$$3x-y=3$$           …… (i)

$$9x-3y=9$$             …… (ii)

From equation (i)-

$$y=3x-3$$          …… (iii)

Substituting (iii) in equation (ii), we get

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

$$9x-9x+9=9$$

$$9=9$$

For all $$x$$ and $$y$$.

Therefore, the given equations have infinite solutions. One of the solution is $$x=1,y=0$$.

(iv) $$0.2x-0.3y=1.3;0.4x+0.5y=2.3$$

Ans (iv): The given equations are-

$$0.2x-0.3y=1.3$$           …… (i)

$$0.4x+0.5y=2.3$$             …… (ii)

From equation (i)-

$$x=\frac{1.3-0.3y}{0.2}$$          …… (iii)

Substituting (iii) in equation (ii), we get

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

$$2.6-0.6y+0.5y=2.3$$

$$2.6-2.3=0.1y$$

$$y=3$$                  …… (iv)

Substituting (iv) in (iii), we get

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

$$x=2$$

Therefore, $$x=2$$ and $$y=3$$.

(v) $$\sqrt{2}x-\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0$$

Ans (v): The given equations are-

$$\sqrt{2}x-\sqrt{3}y=0$$           …… (i)

$$\sqrt{3}x-\sqrt{8}y=0$$             …… (ii)

From equation (i)-

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

Substituting (iii) in equation (ii), we get

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

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

$$y(\frac{-\sqrt{3}}{\sqrt{2}}-2\sqrt{2})=0$$

$$y=0$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=0$$

Therefore, $$x=0$$ and $$y=0$$.

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

Ans (vi): The given equations are-

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

$$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$$             …… (ii)

From equation (i)-

$$x=\frac{-12+10y}{9}$$          …… (iii)

Substituting (iii) in equation (ii), we get

$$\frac{\left(\frac{-12+10y}{9}\right)}{3}+\frac{y}{2}=\frac{13}{6}$$

$$\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}$$

$$\frac{-24+20y+27y}{54}=\frac{13}{6}$$

$$47y=141$$

$$y=3$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=2$$

Therefore, $$x=0$$ and $$y=3$$.

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

Ans: The given equations are-

$$2x+3y=11$$           …… (i)

$$2x-4y=-24$$             …… (ii)

From equation (i)-

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

Substituting (iii) in equation (ii), we get

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

$$11-3y-4y=-24$$

$$-7y=-35$$

$$y=5$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=-2$$

Therefore, $$x=-2$$ and $$y=5$$.

Calculating the value of $$m$$-

$$y=mx+3$$

$$5=-2m+3$$

$$m=-1$$

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

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

Ans (i): Assuming one number be $$x$$ and another number be $$y$$ such that $$y>x$$,

Writing the algebraic representation using the information given in the question-

$$y=3x$$             …… (i)

$$y-x=26$$    …… (ii)

Substituting the value of $$y$$ from equation (i) in equation (ii), we get

$$3x-x=26$$

$$2x=26$$

$$x=13$$                  …… (iii)

Substituting (iii) in (i), we get

$$y=39$$

Therefore, $$x=13$$ and $$y=39$$.

(ii) The larger of two supplementary angles exceeds the smaller by $$\mathbf{18}$$ degrees. Find them.

Ans (ii): Assuming the larger angle be $$x$$ and smaller angle be $$y$$. 

The sum of a pair of supplementary angles is always $${180}^{\circ }$$.

Writing the algebraic representation using the information given in the question-

$$x+y=180$$             …… (i)

$$x-y=18$$    …… (ii)

Substituting the value of $$x$$ from equation (i) in equation (ii), we get

$$180-y-y=18$$

$$162=2y$$

$$y=81$$                  …… (iii)

Substituting (iii) in (i), we get

$$x=99$$

Therefore, the two angles are  $$x={99}^{\circ }$$ and $$y={81}^{\circ }$$.

(iii) The coach of a cricket team buys $$\mathbf{7}$$ bats and 6 balls for $$\mathbf{Rs}\mathbf{3800}$$. Later, she buys $$\mathbf{3}$$ bats and $$\mathbf{5}$$ balls for $$\mathbf{Rs}\mathbf{1750}$$. Find the cost of each bat and each ball.

Ans (iii): Assuming the cost of a bat is $$x$$ and the cost of a ball is $$y$$.

Writing the algebraic representation using the information given in the question-

$$7x+6y=3800$$             …… (i)

$$3x+5y=1750$$    …… (ii)

From equation (i)-

$$y=\frac{3800-7x}{6}$$       …… (iii)

Substituting (iii) in equation (ii)- 

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

$$3x+\frac{9500}{3}-\frac{35x}{6}=1750$$

$$3x-\frac{35x}{6}=1750-\frac{9500}{3}$$

$$\frac{18x-35x}{6}=\frac{5250-9500}{3}$$

$$\frac{17x}{6}=\frac{-4250}{3}$$

$$x=500$$                  …… (iv)

Substituting (iv) in (iii), we get

$$y=\frac{3800-7\left(500\right)}{6}$$

$$y=50$$

Therefore, the bat costs $$Rs500$$ and the ball costs $$Rs50$$.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $$\mathbf{10}\mathbf{km}$$, the charge paid is $$\mathbf{Rs}\mathbf{105}$$ and for a journey of $$\mathbf{15}\mathbf{km}$$, the charge paid is $$\mathbf{Rs}\mathbf{155}$$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $$\mathbf{25}\mathbf{km}$$.

Ans (iv): Assuming the fixed charge be $$Rsx$$ and the per km charge be $$Rs\text{ y}$$.

Writing the algebraic representation using the information given in the question-

$$x+10y=105$$             …… (i)

$$x+15y=155$$    …… (ii)

From equation (i)-

$$x=105-10y$$       …… (iii)

Substituting (iii) in equation (ii)- 

$$105-10y+15y=155$$

$$5y=50$$

$$y=10$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=105-10\left(10\right)$$

$$x=5$$

Therefore, the fixed charge is $$Rs5$$ and the per km charge is $$Rs\text{ 10}$$.

So, charge for $$25km$$ will be-

$$=Rs(x+25y)$$

$$=Rs\text{ 255}$$

(v) A fraction becomes $$\frac{9}{11}$$, if $$\mathbf{2}$$ is added to both the numerator and the denominator. If, $$\mathbf{3}$$ is added to both the numerator and the denominator it becomes $$\frac{5}{6}$$. Find the fraction.

Ans (v): Assuming the fraction be $$\frac{x}{y}$$.

Writing the algebraic representation using the information given in the question-

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

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

$$11x-9y=-4$$             …… (i)

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

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

$$6x-5y=-3$$    …… (ii)

From equation (i)-

$$x=\frac{-4+9y}{11}$$       …… (iii)

Substituting (iii) in equation (ii)- 

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

$$-24+54y-55y=-33$$

$$y=9$$                  …… (iv)

Substituting (iv) in (iii), we get

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

$$x=7$$

Therefore, the fraction is $$\frac{7}{9}$$.

(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?

Ans (vi): Assuming the age of Jacob be $$x$$ and the age of his son be $$y$$.

Writing the algebraic representation using the information given in the .question-

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

$$x-3y=10$$             …… (i)

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

$$x-7y=-30$$    …… (ii)

From equation (i)-

$$x=3y+10$$       …… (iii)

Substituting (iii) in equation (ii)- 

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

$$-4y=-40$$

$$y=10$$                  …… (iv)

Substituting (iv) in (iii), we get

$$x=3\left(10\right)+10$$

$$x=40$$

Therefore, Jacob’s present age is $$40$$ years and his son’s present age is $$10$$ years.

Exercise 3.3

1. Solve the following pair of linear equations by the elimination method and the substitution method-

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

Ans (i): 

Elimination method

The given equations are-

$$x+y=5$$          …… (i)

$$2x-3y=4$$   …… (ii)

Multiplying equation (ii) by $$2$$, we get

$$2x+2y=10$$   …… (iii)

Subtracting equation (ii) from equation (iii), we obtain

$$5y=6$$

$$y=\frac{6}{5}$$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$$x=5-\frac{6}{5}$$

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

Therefore, $$x=\frac{19}{5}$$ and $$y=\frac{6}{5}$$.

Substitution method-

From equation (i) we get

$$x=5-y$$      …… (v)

Substituting (v) in equation (ii), we get

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

$$-5y=-6$$

$$y=\frac{6}{5}$$            …… (vi)

Substituting (vi) in equation (v), we obtain

$$x=5-\frac{6}{5}$$

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

Therefore, $$x=\frac{19}{5}$$ and $$y=\frac{6}{5}$$.

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

Ans (ii): Elimination method

The given equations are-

$$3x+4y=10$$          …… (i)

$$2x-2y=2$$   …… (ii)

Multiplying equation (ii) by $$2$$, we get

$$4x-4y=4$$   …… (iii)

Adding equation (ii) and (iii), we obtain

$$7x=14$$

$$x=2$$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$$6+4y=10$$

$$4y=4$$

$$y=1$$

Therefore, $$x=2$$ and $$y=1$$.

Substitution method-

From equation (ii) we get

$$x=1+y$$      …… (v)

Substituting (v) in equation (i), we get

$$3(1+y)+4y=10$$

$$7y=7$$

$$y=1$$            …… (vi)

Substituting (vi) in equation (v), we obtain

$$x=1+1$$

$$x=2$$

Therefore, $$x=2$$ and $$y=1$$.

(iii) $$\mathbf{3}\mathbf{x}\mathbf{5}\mathbf{y}\mathbf{4}=\mathbf{0}$$ and $$\mathbf{9}\mathbf{x}=\mathbf{2}\mathbf{y}+\mathbf{7}$$

Ans (iii): Elimination method

The given equations are-

$$3x-5y-4=0$$          …… (i)

$$9x=2y+7$$

$$9x-2y=7$$   …… (ii)

Multiplying equation (i) by $$3$$, we get

$$9x-15y-12=0$$   …… (iii)

Subtracting equation (iii) from equation (ii), we obtain

$$13y=-5$$

$$y=-\frac{5}{13}$$              …… (iv)

Substituting the value of (iv) in equation (i), we get

$$3x+\frac{25}{13}-4=0$$

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

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

Therefore, $$x=\frac{9}{13}$$ and $$y=-\frac{5}{13}$$.

Substitution method-

From equation (i) we get

$$x=\frac{5y+4}{3}$$      …… (v)

Substituting (v) in equation (ii), we get

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

$$13y=-5$$

$$y=\frac{-5}{13}$$            …… (vi)

Substituting (vi) in equation (v), we obtain

$$x=\frac{5\left(\frac{-5}{13}\right)+4}{3}$$

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

Therefore, $$x=\frac{9}{13}$$ and $$y=\frac{-5}{13}$$.

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

Ans (iv): Elimination method

The given equations are-

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

$$3x+4y=-6$$          …… (i)

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

$$3x-y=9$$   …… (ii)

Subtracting equation (ii) from equation (i), we obtain

$$5y=-15$$

$$y=-3$$              …… (iv)