Question

Solve the following systems of equations.
$7^x-16y=0,4^x-49y=0$

Answer 1

Let $7^x-16y=\mathrm{0...}\left(1\right)$

$4^x+49y=\mathrm{0...}\left(2\right)$

by eq (1) & (2)

$7^x=16y...\left(3\right)$

$4^x=49y...\left(4\right)$

eq (3) & eq (4)

$\frac{7^x}{4^x}=\frac{{16}^y}{{49}^y}$

$\frac{7^x}{4^x}=\frac{4^2}{7^2}$

${\left(\frac{7}{4}\right)}^x={\left(\frac{4}{7}\right)}^2$

${\left(\frac{7}{4}\right)}^x={\left(\frac{7}{4}\right)}^{-2}$

$\Rightarrow x=-2$

putting value of x in (1)

$7^{-2}=16y$

$\frac{1}{49}=16y$

$y=\frac{1}{49\times 16}=\frac{1}{784}$

Hence $x=-2$

$y=\frac{1}{784}$ sol.