Solve the System of Equations y=x^2+2x-16 y=x+4

$y = x^{2} + 2 x - 16$ $y = x + 4$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} + 2 x - 16 = x + 4$

Step 2

Solve $x^{2} + 2 x - 16 = x + 4$ for $x$ .

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

Subtract $x$ from both sides of the equation.

$x^{2} + 2 x - 16 - x = 4$

Subtract $x$ from $2 x$ .

$x^{2} + x - 16 = 4$

Subtract $4$ from both sides of the equation.

$x^{2} + x - 16 - 4 = 0$

Subtract $4$ from $- 16$ .

$x^{2} + x - 20 = 0$

Factor $x^{2} + x - 20$ using the AC method.

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 $- 20$ and whose sum is $1$ .

$- 4, 5 + y = x + 4$

Write the factored form using these integers.

$(x - 4) (x + 5) = 0$

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

$x - 4 = 0$

$x + 5 = 0 + y = x + 4$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 4) (x + 5) = 0$ true.

$x = 4, - 5$

Step 3

Evaluate $y$ when $x = 4$ .

Substitute $4$ for $x$ .

$y = (4) + 4$

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

Remove parentheses.

$y = 4 + 4$

Remove parentheses.

$y = (4) + 4$

Add $4$ and $4$ .

$y = 8$

Step 4

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

Substitute $- 5$ for $x$ .

$y = (- 5) + 4$

Substitute $- 5$ for $x$ in $y = (- 5) + 4$ and solve for $y$ .

Remove parentheses.

$y = - 5 + 4$

Remove parentheses.

$y = (- 5) + 4$

Add $- 5$ and $4$ .

$y = - 1$

Step 5

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

$(4, 8)$

$(- 5, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 8), (- 5, - 1)$

Equation Form:

$x = 4, y = 8$

$x = - 5, y = - 1$

Step 7

Solve the System of Equations y=x^2+2x-16 y=x+4

Related Questions

Support LCMGCF.com by disabling your adblocker.