Precalculus Examples

$x^{2} + y = 9$ , $x - y + 3 = 0$

Step 1

Subtract $x^{2}$ from both sides of the equation.

$y = 9 - x^{2}$

$x - y + 3 = 0$

Step 2

Replace all occurrences of $y$ with $9 - x^{2}$ in each equation.

Tap for more steps...

Step 2.1

Replace all occurrences of $y$ in $x - y + 3 = 0$ with $9 - x^{2}$ .

$x - (9 - x^{2}) + 3 = 0$

$y = 9 - x^{2}$

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Simplify $x - (9 - x^{2}) + 3$ .

Tap for more steps...

Step 2.2.1.1

Simplify each term.

Tap for more steps...

Step 2.2.1.1.1

Apply the distributive property.

$x - 1 \cdot 9 + x^{2} + 3 = 0$

$y = 9 - x^{2}$

Step 2.2.1.1.2

Multiply $- 1$ by $9$ .

$x - 9 + x^{2} + 3 = 0$

$y = 9 - x^{2}$

Step 2.2.1.1.3

Multiply $- - x^{2}$ .

Tap for more steps...

Step 2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$x - 9 + 1 x^{2} + 3 = 0$

$y = 9 - x^{2}$

Step 2.2.1.1.3.2

Multiply $x^{2}$ by $1$ .

$x - 9 + x^{2} + 3 = 0$

$y = 9 - x^{2}$

$x - 9 + x^{2} + 3 = 0$

$y = 9 - x^{2}$

$x - 9 + x^{2} + 3 = 0$

$y = 9 - x^{2}$

Step 2.2.1.2

Add $- 9$ and $3$ .

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

$y = 9 - x^{2}$

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

$y = 9 - x^{2}$

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

$y = 9 - x^{2}$

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

$y = 9 - x^{2}$

Step 3

Solve for $x$ in $x + x^{2} - 6 = 0$ .

Tap for more steps...

Step 3.1

Factor the left side of the equation.

Tap for more steps...

Step 3.1.1

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

$u + u^{2} - 6 = 0$

$y = 9 - x^{2}$

Step 3.1.2

Factor $u + u^{2} - 6$ using the AC method.

Tap for more steps...

Step 3.1.2.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

$y = 9 - x^{2}$

Step 3.1.2.2

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

$y = 9 - x^{2}$

$(u - 2) (u + 3) = 0$

$y = 9 - x^{2}$

Step 3.1.3

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

$(x - 2) (x + 3) = 0$

$y = 9 - x^{2}$

$(x - 2) (x + 3) = 0$

$y = 9 - x^{2}$

Step 3.2

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

$x - 2 = 0$

$x + 3 = 0$

$y = 9 - x^{2}$

Step 3.3

Set $x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = 9 - x^{2}$

Step 3.3.2

Add $2$ to both sides of the equation.

$x = 2$

$y = 9 - x^{2}$

$x = 2$

$y = 9 - x^{2}$

Step 3.4

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

Tap for more steps...

Step 3.4.1

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

$x + 3 = 0$

$y = 9 - x^{2}$

Step 3.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

$y = 9 - x^{2}$

$x = - 3$

$y = 9 - x^{2}$

Step 3.5

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

$y = 9 - x^{2}$

$x = 2, - 3$

$y = 9 - x^{2}$

Step 4

Replace all occurrences of $x$ with $2$ in each equation.

Tap for more steps...

Step 4.1

Replace all occurrences of $x$ in $y = 9 - x^{2}$ with $2$ .

$y = 9 - {(2)}^{2}$

$x = 2$

Step 4.2

Simplify the right side.

Tap for more steps...

Step 4.2.1

Simplify $9 - {(2)}^{2}$ .

Tap for more steps...

Step 4.2.1.1

Simplify each term.

Tap for more steps...

Step 4.2.1.1.1

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

$y = 9 - 1 \cdot 4$

$x = 2$

Step 4.2.1.1.2

Multiply $- 1$ by $4$ .

$y = 9 - 4$

$x = 2$

$y = 9 - 4$

$x = 2$

Step 4.2.1.2

Subtract $4$ from $9$ .

$y = 5$

$x = 2$

$y = 5$

$x = 2$

$y = 5$

$x = 2$

$y = 5$

$x = 2$

Step 5

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

Tap for more steps...

Step 5.1

Replace all occurrences of $x$ in $y = 9 - x^{2}$ with $- 3$ .

$y = 9 - {(- 3)}^{2}$

$x = - 3$

Step 5.2

Simplify the right side.

Tap for more steps...

Step 5.2.1

Simplify $9 - {(- 3)}^{2}$ .

Tap for more steps...

Step 5.2.1.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1.1

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

$y = 9 - 1 \cdot 9$

$x = - 3$

Step 5.2.1.1.2

Multiply $- 1$ by $9$ .

$y = 9 - 9$

$x = - 3$

$y = 9 - 9$

$x = - 3$

Step 5.2.1.2

Subtract $9$ from $9$ .

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

$y = 0$

$x = - 3$

Step 6

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

$(2, 5)$

$(- 3, 0)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 5), (- 3, 0)$

Equation Form:

$x = 2, y = 5$

$x = - 3, y = 0$

Step 8