NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.2

Class 10 Ex 3.2

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

(i) \[x+y=14;x-y=4\]

Ans: 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

\[\left( 14-y \right)-y=4\]

\[14-2y=4\]

\[10=2y\]

\[y=5\]                  …… (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: 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: 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

\[9x-3\left( 3x-3 \right)=9\]

\[9x-9x+9=9\]

\[9=9\]

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: 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: 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

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

$(\dfrac{-3} { \sqrt{2}}) y-\sqrt{8} y=0$

\[y=0\]                  …… (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: 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, \[x=2\] and \[y=3\].

2. Solve \[\mathbf{2x}+\mathbf{3y}=\mathbf{11}\] and \[\mathbf{2x}-\mathbf{4y}=-\mathbf{24}\] 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: 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, numbers are $13$ & $39$.

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

Ans: 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 \[\mathbf{Rs}\text{ }\mathbf{3800}\]. Later, she buys \[\mathbf{3}\] bats and \[\mathbf{5}\] balls for \[\mathbf{Rs}\text{ }\mathbf{1750}\]. Find the cost of each bat and each ball.

Ans: 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 \[Rs\text{ }500\] and the ball costs \[Rs\text{ }50\].

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

Ans: Assuming the fixed charge be \[Rs\text{ }x\] 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 \[Rs\text{ }5\] and the per km charge is \[Rs\text{ 10}\].

So, charge for \[25\text{ }km\] will be:

\[=Rs\text{ }\left( x+25y \right)\]

\[=Rs\text{ 255}\] 

(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: 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: 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:

\[\left( x+5 \right)=3\left( y+5 \right)\]

\[x-3y=10\]             …… (i)

\[\left( x-5 \right)=7\left( y-5 \right)\]

\[x-7y=-30\]    …… (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.

Conclusion

NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2 act as your perfect guide to conquering linear equations with two variables. This exercise is essential for mastering this fundamental concept. The key understanding from class 10th exercise 3.2 is to make use of methods like elimination and substitution to solve these equations and interpret solutions based on how the equations' graph lines intersect, run parallel, or coincide. By diligently practicing these methods with NCERT Solutions for Class 10 Maths Chapter 3.2 Students can score good marks in their exams.

Class 10 Maths Chapter 3: Exercises Breakdown

CBSE Class 10 Maths Chapter 3 Other Study Materials

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