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: $4x - y + 7= 0$

Question

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: $4x - y + 7= 0$

Team Careers360 · Accepted Answer

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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Question : The graphs of the equations 4x+13y=83 and 12x+34y+52=0 intersect at a point P. The point P also lies on the graph of the equation:Option 1: x+2y−5=0Option 2: 3x−y−7=0Option 3: x−3y−12=0Option 4: 4x−y+7=0

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