Question

In each of 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:
$5x-(5-x)=\frac{1}{2}(3-x);4-3y=\frac{4+y}{3}$.

Answer 1

First equation:
$5x-(5-x)=$ $\frac{1}{2}$$(3-x)$
$5x-5+x=$ $\frac{1}{2}$$(3-x)$

$6x-5=$ $\frac{1}{2}$$(3-x)$

$12x-10=3-x$

$\Rightarrow 13x=13$

$\Rightarrow x=1$

Second equation:

$4-3y=$$\frac{4+y}{3}$
$\Rightarrow 12-9y=4+y$
$\Rightarrow y=$$\frac{4}{5}$
Hence, required point $=(x,y)=(1,\frac{4}{5})$