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Algebra of Derivatives

In eq x2+y=...

Question

In eq $x^{2} + y = 6 x - 14$
if $x = 0, 1, 2$
then the values of $y$ are

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Solution

Consider given the equation,

$x^{2} + y = 6 x - 14$

$x^{2} - 6 x + y = - 14$

Put, $x = 0$

$0^{2} - 6 \times 0 + y = - 14$

$y = - 14$

Again put, x=1

$1^{2} - 6 \times 1 + y = - 14$

$y = - 14 + 6 - 1 = - 9$

Again put, x=2

$2^{2} - 6 \times 2 + y = - 14$
$y = - 2 - 4$

$y = - 6$

Hence, this is the answer.

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