Algebra Examples

$y = x^{2} - 24$ $y = x - 12$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 24 = x - 12$

Step 2

Solve $x^{2} - 24 = x - 12$ for $x$ .

Tap for more steps...

Step 2.1

Subtract $x$ from both sides of the equation.

$x^{2} - 24 - x = - 12$

Step 2.2

Add $12$ to both sides of the equation.

$x^{2} - 24 - x + 12 = 0$

Step 2.3

Add $- 24$ and $12$ .

$x^{2} - x - 12 = 0$

Step 2.4

Factor $x^{2} - x - 12$ using the AC method.

Tap for more steps...

Step 2.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3 + y = x - 12$

Step 2.4.2

Write the factored form using these integers.

$(x - 4) (x + 3) = 0$

Step 2.5

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

$x - 4 = 0$

$x + 3 = 0 + y = x - 12$

Step 2.6

Set $x - 4$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.6.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.7

Set $x + 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.7.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.8

The final solution is all the values that make $(x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Step 3

Evaluate $y$ when $x = 4$ .

Tap for more steps...

Step 3.1

Substitute $4$ for $x$ .

$y = (4) - 12$

Step 3.2

Substitute $4$ for $x$ in $y = (4) - 12$ and solve for $y$ .

Tap for more steps...

Step 3.2.1

Remove parentheses.

$y = 4 - 12$

Step 3.2.2

Remove parentheses.

$y = (4) - 12$

Step 3.2.3

Subtract $12$ from $4$ .

$y = - 8$

Step 4

Evaluate $y$ when $x = - 3$ .

Tap for more steps...

Step 4.1

Substitute $- 3$ for $x$ .

$y = (- 3) - 12$

Step 4.2

Substitute $- 3$ for $x$ in $y = (- 3) - 12$ and solve for $y$ .

Tap for more steps...

Step 4.2.1

Remove parentheses.

$y = - 3 - 12$

Step 4.2.2

Remove parentheses.

$y = (- 3) - 12$

Step 4.2.3

Subtract $12$ from $- 3$ .

$y = - 15$

Step 5

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

$(4, - 8)$

$(- 3, - 15)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 8), (- 3, - 15)$

Equation Form:

$x = 4, y = - 8$

$x = - 3, y = - 15$

Step 7