Precalculus Examples

$x + y = 4$ $y = x^{2} - 8$

Step 1

Replace all occurrences of $y$ with $x^{2} - 8$ in each equation.

Tap for more steps...

Step 1.1

Replace all occurrences of $y$ in $x + y = 4$ with $x^{2} - 8$ .

$x + x^{2} - 8 = 4$

$y = x^{2} - 8$

Step 1.2

Simplify the left side.

Tap for more steps...

Step 1.2.1

Remove parentheses.

$x + x^{2} - 8 = 4$

$y = x^{2} - 8$

$x + x^{2} - 8 = 4$

$y = x^{2} - 8$

$x + x^{2} - 8 = 4$

$y = x^{2} - 8$

Step 2

Solve for $x$ in $x + x^{2} - 8 = 4$ .

Tap for more steps...

Step 2.1

Subtract $4$ from both sides of the equation.

$x + x^{2} - 8 - 4 = 0$

$y = x^{2} - 8$

Step 2.2

Subtract $4$ from $- 8$ .

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

$y = x^{2} - 8$

Step 2.3

Factor the left side of the equation.

Tap for more steps...

Step 2.3.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u + u^{2} - 12 = 0$

$y = x^{2} - 8$

Step 2.3.2

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

Tap for more steps...

Step 2.3.2.1

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 $1$ .

$- 3, 4$

$y = x^{2} - 8$

Step 2.3.2.2

Write the factored form using these integers.

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

$y = x^{2} - 8$

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

$y = x^{2} - 8$

Step 2.3.3

Replace all occurrences of $u$ with $x$ .

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

$y = x^{2} - 8$

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

$y = x^{2} - 8$

Step 2.4

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

$x - 3 = 0$

$x + 4 = 0$

$y = x^{2} - 8$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$x - 3 = 0$

$y = x^{2} - 8$

Step 2.5.2

Add $3$ to both sides of the equation.

$x = 3$

$y = x^{2} - 8$

$x = 3$

$y = x^{2} - 8$

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 $0$ .

$x + 4 = 0$

$y = x^{2} - 8$

Step 2.6.2

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = x^{2} - 8$

$x = - 4$

$y = x^{2} - 8$

Step 2.7

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

$x = 3, - 4$

$y = x^{2} - 8$

$x = 3, - 4$

$y = x^{2} - 8$

Step 3

Replace all occurrences of $x$ with $3$ in each equation.

Tap for more steps...

Step 3.1

Replace all occurrences of $x$ in $y = x^{2} - 8$ with $3$ .

$y = {(3)}^{2} - 8$

$x = 3$

Step 3.2

Simplify the right side.

Tap for more steps...

Step 3.2.1

Simplify ${(3)}^{2} - 8$ .

Tap for more steps...

Step 3.2.1.1

Raise $3$ to the power of $2$ .

$y = 9 - 8$

$x = 3$

Step 3.2.1.2

Subtract $8$ from $9$ .

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

$y = 1$

$x = 3$

Step 4

Replace all occurrences of $x$ with $- 4$ in each equation.

Tap for more steps...

Step 4.1

Replace all occurrences of $x$ in $y = x^{2} - 8$ with $- 4$ .

$y = {(- 4)}^{2} - 8$

$x = - 4$

Step 4.2

Simplify the right side.

Tap for more steps...

Step 4.2.1

Simplify ${(- 4)}^{2} - 8$ .

Tap for more steps...

Step 4.2.1.1

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

$y = 16 - 8$

$x = - 4$

Step 4.2.1.2

Subtract $8$ from $16$ .

$y = 8$

$x = - 4$

$y = 8$

$x = - 4$

$y = 8$

$x = - 4$

$y = 8$

$x = - 4$

Step 5

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

$(3, 1)$

$(- 4, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 1), (- 4, 8)$

Equation Form:

$x = 3, y = 1$

$x = - 4, y = 8$

Step 7