If $x$ and $y$ are real numbers and satisfy the equation
${(2 x + 3 y - 5)}^{2} {(5 x - y - 4)}^{2} = 0$ find $x$ and $y$ .

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Solution

Given : ${(2 x + 3 y - 5)}^{2} {(5 x - y - 4)}^{2} = 0$
$⟹ (2 x + 3 y - 5)^{2} = 0$ and $(5 x - y - 4)^{2} = 0$
$⟹ 2 x + 3 y - 5 = 0$ and $5 x - y - 4 = 0$
$⟹ 2 x + 3 y = 5$ ...... $(i)$ and $5 x - y = 4$ ....... $(i i)$
Solving above equations we get $x = 1$
Substitute $x = 1$ in equation $(i)$ we get
$2 + 3 y = 5 ⟹ y = 1$
Hence, $x = 1$ and $y = 1$

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