Algebra Examples

$y = 3 x + 6$ $y = {(x + 4)}^{2} - 10$

Step 1

Eliminate the equal sides of each equation and combine.

$3 x + 6 = {(x + 4)}^{2} - 10$

Step 2

Solve $3 x + 6 = {(x + 4)}^{2} - 10$ for $x$ .

Tap for more steps...

Step 2.1

Simplify ${(x + 4)}^{2} - 10$ .

Tap for more steps...

Step 2.1.1

Simplify each term.

Tap for more steps...

Step 2.1.1.1

Rewrite ${(x + 4)}^{2}$ as $(x + 4) (x + 4)$ .

$3 x + 6 = (x + 4) (x + 4) - 10$

Step 2.1.1.2

Expand $(x + 4) (x + 4)$ using the FOIL Method.

Tap for more steps...

Step 2.1.1.2.1

Apply the distributive property.

$3 x + 6 = x (x + 4) + 4 (x + 4) - 10$

Step 2.1.1.2.2

Apply the distributive property.

$3 x + 6 = x \cdot x + x \cdot 4 + 4 (x + 4) - 10$

Step 2.1.1.2.3

Apply the distributive property.

$3 x + 6 = x \cdot x + x \cdot 4 + 4 x + 4 \cdot 4 - 10$

Step 2.1.1.3

Simplify and combine like terms.

Tap for more steps...

Step 2.1.1.3.1

Simplify each term.

Tap for more steps...

Step 2.1.1.3.1.1

Multiply $x$ by $x$ .

$3 x + 6 = x^{2} + x \cdot 4 + 4 x + 4 \cdot 4 - 10$

Step 2.1.1.3.1.2

Move $4$ to the left of $x$ .

$3 x + 6 = x^{2} + 4 \cdot x + 4 x + 4 \cdot 4 - 10$

Step 2.1.1.3.1.3

Multiply $4$ by $4$ .

$3 x + 6 = x^{2} + 4 x + 4 x + 16 - 10$

Step 2.1.1.3.2

Add $4 x$ and $4 x$ .

$3 x + 6 = x^{2} + 8 x + 16 - 10$

Step 2.1.2

Subtract $10$ from $16$ .

$3 x + 6 = x^{2} + 8 x + 6$

Step 2.2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 8 x + 6 = 3 x + 6$

Step 2.3

Move all terms containing $x$ to the left side of the equation.

Tap for more steps...

Step 2.3.1

Subtract $3 x$ from both sides of the equation.

$x^{2} + 8 x + 6 - 3 x = 6$

Step 2.3.2

Subtract $3 x$ from $8 x$ .

$x^{2} + 5 x + 6 = 6$

Step 2.4

Subtract $6$ from both sides of the equation.

$x^{2} + 5 x + 6 - 6 = 0$

Step 2.5

Combine the opposite terms in $x^{2} + 5 x + 6 - 6$ .

Tap for more steps...

Step 2.5.1

Subtract $6$ from $6$ .

$x^{2} + 5 x + 0 = 0$

Step 2.5.2

Add $x^{2} + 5 x$ and $0$ .

$x^{2} + 5 x = 0$

Step 2.6

Factor $x$ out of $x^{2} + 5 x$ .

Tap for more steps...

Step 2.6.1

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

$x \cdot x + 5 x = 0$

Step 2.6.2

Factor $x$ out of $5 x$ .

$x \cdot x + x \cdot 5 = 0$

Step 2.6.3

Factor $x$ out of $x \cdot x + x \cdot 5$ .

$x (x + 5) = 0$

Step 2.7

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 5 = 0 + y = {(x + 4)}^{2} - 10$

Step 2.8

Set $x$ equal to $0$ .

$x = 0$

Step 2.9

Set $x + 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.9.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.9.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.10

The final solution is all the values that make $x (x + 5) = 0$ true.

$x = 0, - 5$

Step 3

Evaluate $y$ when $x = 0$ .

Tap for more steps...

Step 3.1

Substitute $0$ for $x$ .

$y = {((0) + 4)}^{2} - 10$

Step 3.2

Substitute $0$ for $x$ in $y = {((0) + 4)}^{2} - 10$ and solve for $y$ .

Tap for more steps...

Step 3.2.1

Remove parentheses.

$y = {(0 + 4)}^{2} - 10$

Step 3.2.2

Remove parentheses.

$y = {((0) + 4)}^{2} - 10$

Step 3.2.3

Simplify ${((0) + 4)}^{2} - 10$ .

Tap for more steps...

Step 3.2.3.1

Simplify each term.

Tap for more steps...

Step 3.2.3.1.1

Add $0$ and $4$ .

$y = 4^{2} - 10$

Step 3.2.3.1.2

Raise $4$ to the power of $2$ .

$y = 16 - 10$

Step 3.2.3.2

Subtract $10$ from $16$ .

$y = 6$

Step 4

Evaluate $y$ when $x = - 5$ .

Tap for more steps...

Step 4.1

Substitute $- 5$ for $x$ .

$y = {((- 5) + 4)}^{2} - 10$

Step 4.2

Substitute $- 5$ for $x$ in $y = {((- 5) + 4)}^{2} - 10$ and solve for $y$ .

Tap for more steps...

Step 4.2.1

Remove parentheses.

$y = {(- 5 + 4)}^{2} - 10$

Step 4.2.2

Remove parentheses.

$y = {((- 5) + 4)}^{2} - 10$

Step 4.2.3

Simplify ${((- 5) + 4)}^{2} - 10$ .

Tap for more steps...

Step 4.2.3.1

Simplify each term.

Tap for more steps...

Step 4.2.3.1.1

Add $- 5$ and $4$ .

$y = {(- 1)}^{2} - 10$

Step 4.2.3.1.2

Raise $- 1$ to the power of $2$ .

$y = 1 - 10$

Step 4.2.3.2

Subtract $10$ from $1$ .

$y = - 9$

Step 5

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

$(0, 6)$

$(- 5, - 9)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(0, 6), (- 5, - 9)$

Equation Form:

$x = 0, y = 6$

$x = - 5, y = - 9$

Step 7