Question

if X =1 y=2is a solution of the equation a
${a}^{2} x + ay = 3$
then find the value of a

Answer 1

A^2.1+a.2=3
A^2+2a-3=0
a^2+3a-a-3=0
a(a+3)-1(a+3)=0
(a+3)(a-1)=0

Either,
(a+3)=0
a=-3

Or,
(a-1)=0
a=1

Answer 2

✍✍HERE IS YOUR SOLTION…⬇⬇
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$\underline{SOLUTION}$ ➡

$Given ,\: \\ \\ x \: = 1 \: \: \: y = 2 \: \: \: is \: \: \: a \: \: solution \: \: of \: \: the \: \: \\ equation \: \: a {}^{2} x +a y = 3$

$Putting \: \: \: the \: \: \: values\: \: of \: \: x \: \: \: and \: \: \: y \: \: \: in \: \:\\ the \: \: \: equation \: : - \: \\ \\ We \: \: get ,\\ \\ a {}^{2} x +a y = 3 \\ \\ = > a {}^{2} \times( 1) + a \times (2) = 3 \\ \\ = > a {}^{2} + 2a - 3 = 0 \\ \\ = > a {}^{2} + 3a - a - 3 = 0 \\ \\ = > a(a + 3) - 1(a + 3 ) = 0 \\ \\ = > (a - 1)(a + 3) = 0 \\ \\ \\ Therefore , \\ \\ a - 1 = 0 \\ \\ = > a = 1 \\ \\ Or, \\ \\ a + 3 = 0 \\ \\ = > a = - 3 \\ \\ So , \\ \\ The \: \: values \: \: of \: \: \bold{ a }\: \: are \: : \: \: \: \boxed{1 ,\: \: -3}$

$\underline{ANSWER}$ ➡ $\boxed{\bold{1 \: , \: -3}}$

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