Algebra Examples

$y = x^{2}$ and $y = - 2 x - 1$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} = - 2 x - 1$

Step 2

Solve $x^{2} = - 2 x - 1$ for $x$ .

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Step 2.1

Add $2 x$ to both sides of the equation.

$x^{2} + 2 x = - 1$

Step 2.2

Add $1$ to both sides of the equation.

$x^{2} + 2 x + 1 = 0$

Step 2.3

Factor using the perfect square rule.

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Step 2.3.1

Rewrite $1$ as $1^{2}$ .

$x^{2} + 2 x + 1^{2} = 0$

Step 2.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 2.3.3

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Step 2.3.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Step 2.4

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Evaluate $y$ when $x = - 1$ .

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Step 3.1

Substitute $- 1$ for $x$ .

$y = - 2 \cdot - 1 - 1$

Step 3.2

Substitute $- 1$ for $x$ in $y = - 2 \cdot - 1 - 1$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = - 2 \cdot - 1 - 1$

Step 3.2.2

Simplify $- 2 \cdot - 1 - 1$ .

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Step 3.2.2.1

Multiply $- 2$ by $- 1$ .

$y = 2 - 1$

Step 3.2.2.2

Subtract $1$ from $2$ .

$y = 1$

Step 4

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 1, 1)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 1, 1)$

Equation Form:

$x = - 1, y = 1$

Step 6