Question

In the following, the co-ordinates of the point $P$ is such that the abscissa is the solution of the first equation and ordinate is the solution of the second equation in $3-2x=7;2y+1=10-2\frac{1}{2}y$. The coordinates obtained can be represented as $(-2,m)$. Then, $m$ is

Answer 1

First equation:
$3-2x=7$
$=>2x=3-7=-4$
$=>x=-2$
Second equation:
$2y+1=10-2$$\frac{1}{2}$$y$
$=>2y+1=10-$$\frac{5}{2}$$y$
$=>2y+$$\frac{5}{2}$ $y=10-1$
$=>$$\frac{9}{2}$ $y=9$
$=>y=2$
Hence, required point $=(x,y)=(-2,2)$