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:
$\frac{2a}{3}-1=\frac{a}{2};\frac{15-4b}{7}=\frac{2b-1}{3}$.

Answer 1

First equation:
$\frac{2a}{3}$ $-1=$ $\frac{a}{2}$
$\Rightarrow$ $\frac{2a}{3}$ $-$ $\frac{a}{2}$ $=1$
On solving we get,
$\Rightarrow a=6$
Second equation:
$\frac{15-4b}{7}$ $=$ $\frac{2b-1}{3}$
On cross - multiplying, we get,
$45-12b=14b-7$
$\Rightarrow b=2$
Hence, required point $=(a,b)=(6,2)$