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Algebra Examples

$x^{2} + y = 0$ , $y = 5 x^{2}$ , $(3, 4)$

Rewrite the equation as $0 = {(3)}^{2} + 4$ .

$0 = {(3)}^{2} + 4$

$4 = 5 {(3)}^{2}$

Simplify ${(3)}^{2} + 4$ .

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Simplify each term.

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Remove parentheses.

$0 = 3^{2} + 4$

$4 = 5 {(3)}^{2}$

Raise $3$ to the power of $2$ .

$0 = 9 + 4$

$4 = 5 {(3)}^{2}$

$0 = 9 + 4$

$4 = 5 {(3)}^{2}$

Add $9$ and $4$ .

$0 = 13$

$4 = 5 {(3)}^{2}$

$0 = 13$

$4 = 5 {(3)}^{2}$

Since $0 \neq 13$ , there are no solutions.

No solution

Rewrite the equation as $5 {(3)}^{2} = 4$ .

No solution

Remove parentheses.

No solution

Raise $3$ to the power of $2$ .

No solution

Multiply $5$ by $9$ .

No solution

Since $45 \neq 4$ , there are no solutions.

No solution

The ordered pair is not a solution to the system of equations.

$(3, 4)$ is not a solution

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$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$