Algebra Examples

$x^{3} + 2 y^{2} + x < 2$ , $(0, 0)$

Insert the values of $x = 0$ and $y = 0$ into the equation to find if the ordered pair is a solution.

${(0)}^{3} + 2 {(0)}^{2} + (0) < 2$

Simplify each term.

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Remove parentheses around $0$ .

$0^{3} + 2 {(0)}^{2} + 0 < 2$

Raising $0$ to any positive power yields $0$ .

$0 + 2 {(0)}^{2} + 0 < 2$

Remove parentheses around $0$ .

$0 + 2 \cdot 0^{2} + 0 < 2$

Raising $0$ to any positive power yields $0$ .

$0 + 2 \cdot 0 + 0 < 2$

Multiply $2$ by $0$ .

$0 + 0 + 0 < 2$

Simplify by adding zeros.

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Add $0$ and $0$ .

$0 + 0 < 2$

Add $0$ and $0$ .

$0 < 2$

Since $0 < 2$ , the equation will always be true.

Always true

Since the equation is always true when the values are used, the ordered pair is a solution.

The ordered pair is a solution to the equation.

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