Mathematics solution NCERT
Class 10 - Chapter 3: Pair of Linear Equations in Two Variables
EXERCISE 3.2
1. Solve the following pair of linear equations by the substitution method.
(i) $x+y=14$ and $x-y=4$
Solution:
From $$ x+y=14 $$ $$ y=14-x $$ Substitute in $$ x-y=4 $$ $$ x-(14-x)=4 $$ $$ 2x-14=4 $$ $$ 2x=18 $$ $$ x=9 $$ Substitute in $$ y=14-9=5 $$Answer:
$$ \boxed{x=9,\qquad y=5} $$(ii) $s-t=3$ and $\dfrac{s}{3}+\dfrac{t}{2}=6$
Solution:
From $$ s-t=3 $$ $$ s=t+3 $$ Substitute into $$ \frac{s}{3}+\frac{t}{2}=6 $$ $$ \frac{t+3}{3}+\frac{t}{2}=6 $$ Multiply throughout by $6$. $$ 2(t+3)+3t=36 $$ $$ 2t+6+3t=36 $$ $$ 5t=30 $$ $$ t=6 $$ Now, $$ s=t+3=9 $$Answer:
$$ \boxed{s=9,\qquad t=6} $$(iii) $3x-y=3$ and $9x-3y=9$
Solution:
From $$ 3x-y=3 $$ $$ y=3x-3 $$ Substitute into $$ 9x-3y=9 $$ $$ 9x-3(3x-3)=9 $$ $$ 9x-9x+9=9 $$ $$ 9=9 $$ This is an identity.Answer:
$$ \boxed{\text{Infinitely many solutions}} $$ The equations represent coincident lines.(iv) $0.2x+0.3y=1.3$ and $0.4x+0.5y=2.3$
Solution:
From $$ 0.2x+0.3y=1.3 $$ $$ 0.2x=1.3-0.3y $$ $$ x=\frac{1.3-0.3y}{0.2} $$ Substitute into $$ 0.4x+0.5y=2.3 $$ $$ 0.4\left(\frac{1.3-0.3y}{0.2}\right)+0.5y=2.3 $$ $$ 2(1.3-0.3y)+0.5y=2.3 $$ $$ 2.6-0.6y+0.5y=2.3 $$ $$ 2.6-0.1y=2.3 $$ $$ -0.1y=-0.3 $$ $$ y=3 $$ Substitute in the first equation. $$ 0.2x+0.9=1.3 $$ $$ 0.2x=0.4 $$ $$ x=2 $$Answer:
$$ \boxed{x=2,\qquad y=3} $$(v) $\sqrt2\,x+\sqrt3\,y=0$ and $\sqrt3\,x-\sqrt8\,y=0$
Solution:
From $$ \sqrt2\,x+\sqrt3\,y=0 $$ $$ \sqrt2\,x=-\sqrt3\,y $$ $$ x=-\frac{\sqrt3}{\sqrt2}y $$ Substitute into $$ \sqrt3\,x-\sqrt8\,y=0 $$ $$ \sqrt3\left(-\frac{\sqrt3}{\sqrt2}y\right)-\sqrt8\,y=0 $$ $$ -\frac{3}{\sqrt2}y-2\sqrt2\,y=0 $$ Multiply by $\sqrt2$. $$ -3y-4y=0 $$ $$ -7y=0 $$ $$ y=0 $$ Therefore, $$ x=0 $$Answer:
$$ \boxed{x=0,\qquad y=0} $$(vi) $\dfrac{3x}{2}-\dfrac{5y}{3}=-2$ and $\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}$
Solution:
From $$ \frac{x}{3}+\frac{y}{2}=\frac{13}{6} $$ Multiply throughout by $6$. $$ 2x+3y=13 $$ $$ 2x=13-3y $$ $$ x=\frac{13-3y}{2} $$ Substitute into $$ \frac{3x}{2}-\frac{5y}{3}=-2 $$ $$ \frac{3}{2}\left(\frac{13-3y}{2}\right)-\frac{5y}{3}=-2 $$ Multiply throughout by $12$. $$ 9(13-3y)-20y=-24 $$ $$ 117-27y-20y=-24 $$ $$ 117-47y=-24 $$ $$ 47y=141 $$ $$ y=3 $$ Substitute into $$ 2x+3y=13 $$ $$ 2x+9=13 $$ $$ 2x=4 $$ $$ x=2 $$Answer:
$$ \boxed{x=2,\qquad y=3} $$2. Solve $2x+3y=11$ and $2x-4y=-24$ and hence find the value of $m$ for which $y=mx+3$.
Solution:
Given, $$ 2x+3y=11 $$ $$ 2x-4y=-24 $$ Subtract the second equation from the first. $$ (2x+3y)-(2x-4y)=11-(-24) $$ $$ 7y=35 $$ $$ y=5 $$ Substitute in $$ 2x+3y=11 $$ $$ 2x+3(5)=11 $$ $$ 2x+15=11 $$ $$ 2x=-4 $$ $$ x=-2 $$ Now use $$ y=mx+3 $$ Substitute $$ x=-2,\qquad y=5 $$ $$ 5=m(-2)+3 $$ $$ 2=-2m $$ $$ m=-1 $$Answer:
$$ \boxed{x=-2,\qquad y=5,\qquad m=-1} $$3. Form the pair of linear equations for the following problems and solve them by substitution method.
(i) The difference between two numbers is $26$ and one number is three times the other. Find them.
Solution:
Let the two numbers be $x$ and $y$. Given, $$ x-y=26 $$ $$ x=3y $$ Substitute $$ 3y-y=26 $$ $$ 2y=26 $$ $$ y=13 $$ $$ x=3(13)=39 $$Answer:
$$ \boxed{\text{Numbers are }39\text{ and }13} $$(ii) The larger of two supplementary angles exceeds the smaller by $18^\circ$. Find them.
Solution:
Let the larger angle be $x^\circ$ and the smaller angle be $y^\circ$. Since they are supplementary, $$ x+y=180 $$ Also, $$ x-y=18 $$ From $$ x=18+y $$ Substitute into $$ x+y=180 $$ $$ 18+y+y=180 $$ $$ 2y=162 $$ $$ y=81 $$ $$ x=99 $$Answer:
$$ \boxed{99^\circ\text{ and }81^\circ} $$(iii) The coach of a cricket team buys $7$ bats and $6$ balls for ₹$3800$. Later, she buys $3$ bats and $5$ balls for ₹$1750$. Find the cost of each bat and each ball.
Solution:
Let the cost of one bat be ₹$x$ and one ball be ₹$y$. Then, $$ 7x+6y=3800 $$ $$ 3x+5y=1750 $$ From $$ 3x=1750-5y $$ $$ x=\frac{1750-5y}{3} $$ Substitute into the first equation. $$ 7\left(\frac{1750-5y}{3}\right)+6y=3800 $$ Multiply by $3$. $$ 12250-35y+18y=11400 $$ $$ 12250-17y=11400 $$ $$ 17y=850 $$ $$ y=50 $$ Now, $$ 3x+5(50)=1750 $$ $$ 3x=1500 $$ $$ x=500 $$Answer:
$$ \boxed{\text{Bat}=₹500,\qquad \text{Ball}=₹50} $$(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10$ km, the charge paid is ₹$105$ and for a journey of $15$ km, the charge paid is ₹$155$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25$ km?
Solution:
Let the fixed charge be ₹$x$ and the charge per km be ₹$y$. Then, $$ x+10y=105 $$ $$ x+15y=155 $$ From $$ x=105-10y $$ Substitute into the second equation. $$ 105-10y+15y=155 $$ $$ 5y=50 $$ $$ y=10 $$ Now, $$ x=105-100=5 $$ Fare for $25$ km, $$ =5+25(10) $$ $$ =255 $$Answer:
$$ \boxed{\text{Fixed charge}=₹5} $$ $$ \boxed{\text{Charge per km}=₹10} $$ $$ \boxed{\text{Fare for }25\text{ km}=₹255} $$(v) A fraction becomes $\dfrac{9}{11}$ if $2$ is added to both the numerator and the denominator. If $3$ is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$. Find the fraction.
Solution:
Let the numerator be $x$ and the denominator be $y$. Then, $$ \frac{x+2}{y+2}=\frac{9}{11} $$ Cross multiply. $$ 11(x+2)=9(y+2) $$ $$ 11x-9y=-4 $$ Also, $$ \frac{x+3}{y+3}=\frac56 $$ Cross multiply. $$ 6(x+3)=5(y+3) $$ $$ 6x-5y=-3 $$ From $$ 6x=5y-3 $$ $$ x=\frac{5y-3}{6} $$ Substitute into $$ 11x-9y=-4 $$ $$ 11\left(\frac{5y-3}{6}\right)-9y=-4 $$ Multiply by $6$. $$ 55y-33-54y=-24 $$ $$ y=9 $$ Now, $$ 6x-5(9)=-3 $$ $$ 6x=42 $$ $$ x=7 $$ Therefore, the required fraction is $$ \boxed{\frac79} $$(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?
Solution:
Let the present age of Jacob be $x$ years and the present age of his son be $y$ years.
Five years hence,
$$ x+5=3(y+5) $$ $$ x+5=3y+15 $$ $$ x=3y+10 $$Five years ago,
$$ x-5=7(y-5) $$ $$ x-5=7y-35 $$ $$ x=7y-30 $$Substitute
$$ x=3y+10 $$into
$$ x=7y-30 $$ $$ 3y+10=7y-30 $$ $$ 40=4y $$ $$ y=10 $$Now,
$$ x=3(10)+10 $$ $$ x=40 $$Verification:
Five years hence,
$$ 45=3\times15 $$Five years ago,
$$ 35=7\times5 $$Both conditions are satisfied.
Answer:
$$ \boxed{\text{Jacob's present age }=40\text{ years}} $$ $$ \boxed{\text{Son's present age }=10\text{ years}} $$