Question

Consider the equation of a line $y-2x=4$ Comment on:

$\left(i\right)x=0,y=4$ is a solution of the equation.

$\left(ii\right)x=1,y=5$ is a solution of the equation

$\left(iii\right)(-4,-4)$ lies on the graph of the equation.

Answer 1

$\left(i\right)x=0,y=4$

L.H.S$=y-2x=4-2\times 0=4-0=4$

So,$x=0,y=4$ is a solution of the equation $y-2x=4$

$\left(ii\right)x=1,y=5$

L.H.S$=y-2x=5-2\times 1=5-2=3\ne$R.H.S

So,$x=1,y=5$ is not a solution of the equation $y-2x=4$

$\left(iii\right)x=-4,y=-4$

L.H.S$=-4-2\times -4=-4+8=4=$R.H.S

So, $(-4,-4)$ lies on the graph of the equation $y-2x=4$