Exercise-3.2, Class 10th, Maths, Chapter-3, NCERT

1. Solve the following pairs by the substitution method.

(i) $x+y=14,x-y=4.$
Add the two equations: $2x=18\Rightarrow x=9.$Then $y=14-9=5.$
Answer: $(x,y)=(9,5)$

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(ii) $s-t=3,{\displaystyle \frac{s}{3}}+{\displaystyle \frac{t}{2}}=6.$
From first: $s=t+3.$

Substitute:

$$\frac{t+3}{3}+\frac{t}{2}=6\Rightarrow \frac{2(t+3)+3t}{6}=6\Rightarrow 2t+6+3t=36\Rightarrow 5t=30\Rightarrow t=6.$$Then $s=6+3=9.$
Answer: $(s,t)=(9,6)$.

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(iii) $3x-y=3,9x-3y=9.$
Observe second equation is $3\times (3x-y)=9$. So the two equations are dependent — they represent the same line.
From $3x-y=3\Rightarrow y=3x-3$. Any $(x,y)$ with $y=3x-3$ is a solution.
Answer: Infinitely many solutions.

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(iv) $0.2x+0.3y=1.3,0.4x+0.5y=2.3$
Multiply both equations by 10 to clear decimals:

$$2x+3y=13,4x+5y=23.$$

From the first: $2x=13-3y\Rightarrow x={\displaystyle \frac{13-3y}{2}}\mathrm{.}$

Substitute into second (or eliminate): multiply first by 2: $4x+6y=26$.

Subtract second:

$$(4x+6y)-(4x+5y)=26-23\Rightarrow y=3.$$

Then $2x+3\cdot 3=13\Rightarrow 2x=4\Rightarrow x=2.$

Answer: $(x,y)=(2,3)$

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(v) $\sqrt{2}x+\sqrt{3}y=0,\sqrt{3}x-\sqrt{8}y=0.$
Note $\sqrt{8}=2\sqrt{2}$. From the first: $x=-{\displaystyle \frac{\sqrt{3}}{\sqrt{2}}}y$.

Substitute in second:

$$\sqrt{3}\text{\hspace{0.17em}⁣}(-\frac{\sqrt{3}}{\sqrt{2}}y)-2\sqrt{2}y=0\Rightarrow -\frac{3}{\sqrt{2}}y-2\sqrt{2}y=0.$$

Multiply by $\sqrt{2}$: $-3y-4y=-7y=0\Rightarrow y=0$
Answer: $(x,y)=(0,0)$

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(vi) ${\displaystyle \frac{3}{2}}x-{\displaystyle \frac{5}{3}}y=-2,{\displaystyle \frac{x}{3}}+{\displaystyle \frac{y}{2}}={\displaystyle \frac{13}{6}}\mathrm{.}$
Clear denominators by multiplying both equations by 6:

$$9x-10y=-12,2x+3y=13.$$

From second: $2x=13-3y\Rightarrow x={\displaystyle \frac{13-3y}{2}}\mathrm{.}$

Substitute or eliminate: multiply second by 5 → $10x+15y=65$Add with $9x-10y=-12$ after appropriate ops or eliminate y: multiply first by 3: $27x-30y=-36$; multiply second by 10: $20x+30y=130$.

Add:

$$47x=94\Rightarrow x=2,2(2)+3y=13\Rightarrow y=3.$$

Answer: $(x,y)=(2,3)$.

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2. Solve $2x+3y=11$ and $2x-4y=-24$. Hence find $m$ for which $y=mx+3$.

Subtract second from first:

$$(2x+3y)-(2x-4y)=11-(-24)\Rightarrow 7y=35\Rightarrow y=5.$$Then $2x+3\cdot 5=11\Rightarrow 2x=-4\Rightarrow x=-2.$
We need $m$ so that $y=mx+3$ holds at $(x,y)=(-2,5)$:

$$5=m(-2)+3\Rightarrow -2m=2\Rightarrow m=-1.$$

Answer: solution $(x,y)=(-2,5)$. Value $m=-1$.

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3. Form the pair of linear equations for each situation and solve by substitution.

(i) The difference between two numbers is 26 and one number is three times the other.
Let smaller $=a$, larger $=b$. Given $b-a=26$ and $b=3a$. Substitute: $3a-a=26\Rightarrow 2a=26\Rightarrow a=13,b=39.$
Answer: Numbers are $13$ and $39$.

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(ii) The larger of two supplementary angles exceeds the smaller by 18°. Find them.
Let smaller $=\theta$, larger $=\varphi$. $\theta +\varphi =180,\varphi -\theta =18.$ Add: $2\varphi =198\Rightarrow \varphi ={99}^{\circ },\theta ={81}^{\circ }\mathrm{.}$
Answer: ${81}^{\circ }$ and ${99}^{\circ }$.

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(iii) Coach buys 7 bats and 6 balls for ₹3800. Later 3 bats and 5 balls for ₹1750. Find cost of bat and ball.
Let bat cost $B$, ball cost $C$:

$$7B+6C=3800,3B+5C=1750.$$

Multiply second by 2: $6B+10C=3500$.

Subtract from first×1 (or eliminate): From $7B+6C=3800$ minus the $6B+10C=3500$

$$B-4C=300\text{(but easier: multiply first by5 and second by6 → eliminate)}\mathrm{.}$$

(Direct elimination:) Multiply first by5: $35B+30C=19000$

Multiply second by6: $18B+30C=10500$

Subtract: $17B=8500\Rightarrow B=500$
$3(500)+5C=1750\Rightarrow 1500+5C=1750\Rightarrow C=50.$
Answer: Bat = ₹500, Ball = ₹50.

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(iv) Taxi: fixed charge + per km charge. For 10 km pay ₹105; for 15 km pay ₹155. Find fixed charge, per km charge. Cost for 25 km?
Let fixed charge$=F$, rate per km $=r$

$$F+10r=105,F+15r=155.$$

Subtract: $5r=50\Rightarrow r=10$.

Then $F=105-10\cdot 10=5$

$F+25r=5+250=255$
Answer: Fixed = ₹5, per km = ₹10, for 25 km = ₹255.

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(v) A fraction becomes $\frac{9}{11}$ if 2 is added to numerator and denominator; becomes $\frac{5}{6}$ if 3 is added to both. Find the fraction.
Let fraction $={\displaystyle \frac{x}{y}}$.

$$\frac{x+2}{y+2}=\frac{9}{11}\Rightarrow 11(x+2)=9(y+2)\Rightarrow 11x-9y=-4.$$
$$\frac{x+3}{y+3}=\frac{5}{6}\Rightarrow 6(x+3)=5(y+3)\Rightarrow 6x-5y=-3.$$

Solve:
Multiply second by11: $66x-55y=-33$. Multiply first by6: $66x-54y=-24$

Subtract: $(-55y)-(-54y)=-y=-9\Rightarrow y=9.$ Then $6x-5(9)=-3\Rightarrow 6x-45=-3\Rightarrow 6x=42\Rightarrow x=7.$
Answer: Fraction is ${\displaystyle \frac{7}{9}}$

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(vi) Five years hence Jacob’s age will be three times his son’s; five years ago Jacob’s age was seven times his son’s. Find present ages.
Let present ages Jacob $=J$ son $=S$

$$J+5=3(S+5)\Rightarrow J-3S=10,$$
$$J-5=7(S-5)\Rightarrow J-7S=-30.$$

Subtract second from first: $(J-3S)-(J-7S)=10-(-30)\Rightarrow 4S=40\Rightarrow S=10.$

Then $J=3S+10=3\cdot 10+10=40.$
Answer: Jacob = 40 years, son = 10 years.