Precalculus Examples

$y = x^{2} - 4 x - 10$ , $y = - x^{2} - 2 x + 14$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 4 x - 10 = - x^{2} - 2 x + 14$

Step 2

Solve $x^{2} - 4 x - 10 = - x^{2} - 2 x + 14$ for $x$ .

Tap for more steps...

Step 2.1

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

Tap for more steps...

Step 2.1.1

Add $x^{2}$ to both sides of the equation.

$x^{2} - 4 x - 10 + x^{2} = - 2 x + 14$

Step 2.1.2

Add $2 x$ to both sides of the equation.

$x^{2} - 4 x - 10 + x^{2} + 2 x = 14$

Step 2.1.3

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} - 4 x - 10 + 2 x = 14$

Step 2.1.4

Add $- 4 x$ and $2 x$ .

$2 x^{2} - 2 x - 10 = 14$

Step 2.2

Subtract $14$ from both sides of the equation.

$2 x^{2} - 2 x - 10 - 14 = 0$

Step 2.3

Subtract $14$ from $- 10$ .

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

Step 2.4

Factor the left side of the equation.

Tap for more steps...

Step 2.4.1

Factor $2$ out of $2 x^{2} - 2 x - 24$ .

Tap for more steps...

Step 2.4.1.1

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

$2 (x^{2}) - 2 x - 24 = 0$

Step 2.4.1.2

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

$2 (x^{2}) + 2 (- x) - 24 = 0$

Step 2.4.1.3

Factor $2$ out of $- 24$ .

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

Step 2.4.1.4

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

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

Step 2.4.1.5

Factor $2$ out of $2 (x^{2} - x) + 2 (- 12)$ .

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

Step 2.4.2

Factor.

Tap for more steps...

Step 2.4.2.1

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

Tap for more steps...

Step 2.4.2.1.1

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 $- 1$ .

$- 4, 3 + y = - x^{2} - 2 x + 14$

Step 2.4.2.1.2

Write the factored form using these integers.

$2 ((x - 4) (x + 3)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$2 (x - 4) (x + 3) = 0$

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 $0$ .

$x - 4 = 0$

$x + 3 = 0 + y = - x^{2} - 2 x + 14$

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 $2 (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Step 3

Evaluate $y$ when $x = 4$ .

Tap for more steps...

Step 3.1

Substitute $4$ for $x$ .

$y = - {(4)}^{2} - 2 \cdot 4 + 14$

Step 3.2

Substitute $4$ for $x$ in $y = - {(4)}^{2} - 2 \cdot 4 + 14$ and solve for $y$ .

Tap for more steps...

Step 3.2.1

Remove parentheses.

$y = - 4^{2} - 2 \cdot 4 + 14$

Step 3.2.2

Remove parentheses.

$y = - {(4)}^{2} - 2 \cdot 4 + 14$

Step 3.2.3

Simplify $- {(4)}^{2} - 2 \cdot 4 + 14$ .

Tap for more steps...

Step 3.2.3.1

Simplify each term.

Tap for more steps...

Step 3.2.3.1.1

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

$y = - 1 \cdot 16 - 2 \cdot 4 + 14$

Step 3.2.3.1.2

Multiply $- 1$ by $16$ .

$y = - 16 - 2 \cdot 4 + 14$

Step 3.2.3.1.3

Multiply $- 2$ by $4$ .

$y = - 16 - 8 + 14$

Step 3.2.3.2

Simplify by adding and subtracting.

Tap for more steps...

Step 3.2.3.2.1

Subtract $8$ from $- 16$ .

$y = - 24 + 14$

Step 3.2.3.2.2

Add $- 24$ and $14$ .

$y = - 10$

Step 4

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

Tap for more steps...

Step 4.1

Substitute $- 3$ for $x$ .

$y = - {(- 3)}^{2} - 2 \cdot - 3 + 14$

Step 4.2

Substitute $- 3$ for $x$ in $y = - {(- 3)}^{2} - 2 \cdot - 3 + 14$ and solve for $y$ .

Tap for more steps...

Step 4.2.1

Remove parentheses.

$y = - {(- 3)}^{2} - 2 \cdot - 3 + 14$

Step 4.2.2

Simplify $- {(- 3)}^{2} - 2 \cdot - 3 + 14$ .

Tap for more steps...

Step 4.2.2.1

Simplify each term.

Tap for more steps...

Step 4.2.2.1.1

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

$y = - 1 \cdot 9 - 2 \cdot - 3 + 14$

Step 4.2.2.1.2

Multiply $- 1$ by $9$ .

$y = - 9 - 2 \cdot - 3 + 14$

Step 4.2.2.1.3

Multiply $- 2$ by $- 3$ .

$y = - 9 + 6 + 14$

Step 4.2.2.2

Simplify by adding numbers.

Tap for more steps...

Step 4.2.2.2.1

Add $- 9$ and $6$ .

$y = - 3 + 14$

Step 4.2.2.2.2

Add $- 3$ and $14$ .

$y = 11$

Step 5

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

$(4, - 10)$

$(- 3, 11)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 10), (- 3, 11)$

Equation Form:

$x = 4, y = - 10$

$x = - 3, y = 11$

Step 7