Algebra Examples

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

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

Tap for more steps...

Step 2.1

Subtract $2 x$ from both sides of the equation.

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

Step 2.2

Subtract $2$ from both sides of the equation.

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

Step 2.3

Subtract $2$ from $1$ .

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

Step 2.4

Factor the left side of the equation.

Tap for more steps...

Step 2.4.1

Factor $- 1$ out of $- x^{2} - 2 x - 1$ .

Tap for more steps...

Step 2.4.1.1

Factor $- 1$ out of $- x^{2}$ .

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

Step 2.4.1.2

Factor $- 1$ out of $- 2 x$ .

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

Step 2.4.1.3

Rewrite $- 1$ as $- 1 (1)$ .

$- (x^{2}) - (2 x) - 1 \cdot 1 = 0$

Step 2.4.1.4

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

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

Step 2.4.1.5

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (1)$ .

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

Step 2.4.2

Factor using the perfect square rule.

Tap for more steps...

Step 2.4.2.1

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

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

Step 2.4.2.2

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 2.4.2.3

Rewrite the polynomial.

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

Step 2.4.2.4

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 2.5

Divide each term in $- {(x + 1)}^{2} = 0$ by $- 1$ and simplify.

Tap for more steps...

Step 2.5.1

Divide each term in $- {(x + 1)}^{2} = 0$ by $- 1$ .

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

Step 2.5.2

Simplify the left side.

Tap for more steps...

Step 2.5.2.1

Dividing two negative values results in a positive value.

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

Step 2.5.2.2

Divide ${(x + 1)}^{2}$ by $1$ .

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

Step 2.5.3

Simplify the right side.

Tap for more steps...

Step 2.5.3.1

Divide $0$ by $- 1$ .

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

Step 2.6

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

$x + 1 = 0$

Step 2.7

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

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

Tap for more steps...

Step 3.1

Substitute $- 1$ for $x$ .

$y = 2 (- 1) + 2$

Step 3.2

Simplify $2 (- 1) + 2$ .

Tap for more steps...

Step 3.2.1

Multiply $2$ by $- 1$ .

$y = - 2 + 2$

Step 3.2.2

Add $- 2$ and $2$ .

$y = 0$

Step 4

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

$(- 1, 0)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 1, 0)$

Equation Form:

$x = - 1, y = 0$

Step 6