Question : The graphs of the equations $4 x+\frac{1}{3} y=\frac{8}{3}$ and $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ intersect at a point P. The point P also lies on the graph of the equation:
Option 1: $x + 2y - 5 = 0$
Option 2: $3x - y - 7 = 0$
Option 3: $x - 3y - 12= 0$
Option 4: $4x - y + 7= 0$
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Correct Answer: $3x - y - 7 = 0$
Solution : $4 x+\frac{1}{3} y = \frac{8}{3}$ ⇒ $12x+y=8$ -------------------(1) $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ ⇒ $2x+3y=-10$ ⇒ $12x+ 18y = -60$ ----------------(2) Subtracting (1) from (2), we get, ⇒ $y = -4$ and $x=1$ So, P(1, –4) Put this value in the equations given in the options: 3$x$ – $y$ – 7 = 0 ⇒ (7 – 7) = 0 ⇒ 0 = 0 it satisfies point P. Hence, the correct answer is $3x-y- 7 = 0$
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