Question

In the following, find the co-ordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation :

$3-2x=7;2y+1=10-2\frac{1}{2}y.$

• A. $(-2,2)$
• B. $(-4,2)$
• C. $(-2,-2)$
• D. $(-4,-2)$

Answer 1

The correct option is A $(-2,2)$

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}$ $y=101$

$=>$$\frac{9}{2}$ $y=9$

$=>y=2$

Hence, required point $=(x,y)=(-2,2)$