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

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

Claim : Solution is $(6, 2)$

If true then enter $1$ and if false then enter $0$

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Solution

First equation:

$\frac{2 a}{3}$ $- 1 =$ $\frac{a}{2}$

$\Rightarrow \frac{2 a}{3}$ $-$ $\frac{a}{2}$ $= 1$

On solving we get,

$\Rightarrow a = 6$
Second equation:

$\frac{15 - 4 b}{7}$ $=$ $\frac{2 b - 1}{3}$

On cross - multiplying, we get,

$45 - 12 b = 14 b - 7$

$\Rightarrow b = 2$
Hence, required point $= (a, b) = (6, 2)$

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