Pair of Linear Equations – Exercise 3.2

(i)
$x + y = 14 \Rightarrow x = 14 - y$ ...(1)
Substitute in $x - y = 4$:
$(14 - y) - y = 4 \Rightarrow 14 - 2y = 4 \Rightarrow 2y = 10 \Rightarrow y = 5$.
Put $y=5$ in (1): $x = 14 - 5 = 9$.
✅ $x = 9, y = 5$

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(ii)
$s - t = 3 \Rightarrow s = t + 3$ ...(1)
Substitute in $\frac{s}{3} + \frac{t}{2} = 6$:
$\frac{t+3}{3} + \frac{t}{2} = 6 \Rightarrow \frac{2(t+3) + 3t}{6} = 6$
$2t + 6 + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6$.
Put $t=6$ in (1): $s = 6 + 3 = 9$.
✅ $s = 9, t = 6$

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(iii)
$3x - y = 3 \Rightarrow y = 3x - 3$ ...(1)
Substitute in $9x - 3y = 9$:
$9x - 3(3x - 3) = 9 \Rightarrow 9x - 9x + 9 = 9 \Rightarrow 9 = 9$.
This statement is true for all values of $x$.
✅ The pair of equations has infinitely many solutions.

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(iv)
Multiply both equations by 10 to remove decimals:
$2x + 3y = 13$ ...(1)
$4x + 5y = 23$ ...(2)
From (1): $x = \frac{13 - 3y}{2}$. Substitute in (2):
$4(\frac{13 - 3y}{2}) + 5y = 23 \Rightarrow 2(13 - 3y) + 5y = 23$
$26 - 6y + 5y = 23 \Rightarrow -y = -3 \Rightarrow y = 3$.
Put $y=3$ in $x$: $x = \frac{13 - 9}{2} = 2$.
✅ $x = 2, y = 3$

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(v)
$\sqrt{2}x + \sqrt{3}y = 0 \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}y$ ...(1)
Substitute in $\sqrt{3}x - \sqrt{8}y = 0$:
$\sqrt{3}(-\frac{\sqrt{3}}{\sqrt{2}}y) - 2\sqrt{2}y = 0 \Rightarrow -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$
$y(-\frac{3}{\sqrt{2}} - 2\sqrt{2}) = 0 \Rightarrow y = 0$.
Put $y=0$ in (1): $x = 0$.
✅ $x = 0, y = 0$

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(vi)
Simplify equations:
$\frac{3x}{2} - \frac{5y}{3} = -2 \Rightarrow 9x - 10y = -12$ ...(1)
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6} \Rightarrow 2x + 3y = 13$ ...(2)
From (2): $x = \frac{13 - 3y}{2}$. Substitute in (1):
$9(\frac{13 - 3y}{2}) - 10y = -12 \Rightarrow \frac{117 - 27y - 20y}{2} = -12$
$117 - 47y = -24 \Rightarrow 47y = 141 \Rightarrow y = 3$.
Put $y=3$ in $x$: $x = \frac{13 - 9}{2} = 2$.
✅ $x = 2, y = 3$

Step 1: Solve for x and y
$2x + 3y = 11$ ...(1)
$2x - 4y = -24$ ...(2)
From (2): $2x = 4y - 24 \Rightarrow x = 2y - 12$.
Substitute in (1):
$2(2y - 12) + 3y = 11 \Rightarrow 4y - 24 + 3y = 11$
$7y = 35 \Rightarrow y = 5$.
Put $y=5$ in $x$: $x = 2(5) - 12 = -2$.
Solution: $x = -2, y = 5$.

Step 2: Find m
$y = mx + 3$
$5 = m(-2) + 3 \Rightarrow 2 = -2m \Rightarrow m = -1$.
✅ $m = -1$

(i) Let numbers be $x$ and $y$ ($x > y$).
$x - y = 26$ ...(1)
$x = 3y$ ...(2)
Substitute (2) in (1): $3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13$.
$x = 3(13) = 39$.
✅ Numbers are 39 and 13.

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(ii) Let angles be $x$ and $y$ ($x > y$).
Supplementary: $x + y = 180^\circ$ ...(1)
Difference: $x = y + 18^\circ$ ...(2)
Substitute (2) in (1): $(y + 18) + y = 180 \Rightarrow 2y = 162 \Rightarrow y = 81^\circ$.
$x = 81 + 18 = 99^\circ$.
✅ Angles are $99^\circ$ and $81^\circ$.

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(iii) Let cost of bat = $x$, ball = $y$.
$7x + 6y = 3800$ ...(1)
$3x + 5y = 1750 \Rightarrow 3x = 1750 - 5y \Rightarrow x = \frac{1750 - 5y}{3}$ ...(2)
Substitute (2) in (1):
$7(\frac{1750 - 5y}{3}) + 6y = 3800 \Rightarrow 12250 - 35y + 18y = 11400$
$-17y = -850 \Rightarrow y = 50$.
Put $y=50$ in (2): $x = \frac{1750 - 250}{3} = 500$.
✅ Cost of bat = ₹ 500, Cost of ball = ₹ 50.

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(iv) Let fixed charge = $x$, charge per km = $y$.
$x + 10y = 105$ ...(1)
$x + 15y = 155$ ...(2)
From (1): $x = 105 - 10y$. Substitute in (2):
$(105 - 10y) + 15y = 155 \Rightarrow 5y = 50 \Rightarrow y = 10$.
$x = 105 - 100 = 5$.
Charge for 25 km = $x + 25y = 5 + 25(10) = 255$.
✅ Fixed charge = ₹ 5, Per km = ₹ 10. Cost for 25 km = ₹ 255.

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(v) Let fraction be $\frac{x}{y}$.
$\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x + 22 = 9y + 18 \Rightarrow 11x - 9y = -4$ ...(1)
$\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x + 18 = 5y + 15 \Rightarrow 6x - 5y = -3$ ...(2)
From (2): $x = \frac{5y - 3}{6}$. Substitute in (1):
$11(\frac{5y - 3}{6}) - 9y = -4 \Rightarrow 55y - 33 - 54y = -24$
$y = 9$.
Put $y=9$ in $x$: $x = \frac{45 - 3}{6} = 7$.
✅ Fraction is $\frac{7}{9}$.

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(vi) Let Jacob's age = $x$, Son's age = $y$.
5 years hence: $x + 5 = 3(y + 5) \Rightarrow x - 3y = 10$ ...(1)
5 years ago: $x - 5 = 7(y - 5) \Rightarrow x - 7y = -30$ ...(2)
From (1): $x = 3y + 10$. Substitute in (2):
$(3y + 10) - 7y = -30 \Rightarrow -4y = -40 \Rightarrow y = 10$.
$x = 3(10) + 10 = 40$.
✅ Jacob is 40 years old, Son is 10 years old.