Solve by Substitution x+y=4 , y=x^2-16

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

Step 1

Replace all occurrences of $y$ with $x^{2} - 16$ in each equation.

Replace all occurrences of $y$ in $x + y = 4$ with $x^{2} - 16$ .

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

$y = x^{2} - 16$

Simplify the left side.

Remove parentheses.

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

$y = x^{2} - 16$

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

$y = x^{2} - 16$

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

$y = x^{2} - 16$

Step 2

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

Subtract $4$ from both sides of the equation.

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

$y = x^{2} - 16$

Subtract $4$ from $- 16$ .

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

$y = x^{2} - 16$

Factor the left side of the equation.

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

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

$y = x^{2} - 16$

Factor $u + u^{2} - 20$ 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 $1$ .

$- 4, 5$

$y = x^{2} - 16$

Write the factored form using these integers.

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

$y = x^{2} - 16$

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

$y = x^{2} - 16$

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

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

$y = x^{2} - 16$

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

$y = x^{2} - 16$

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

$x - 4 = 0$

$x + 5 = 0$

$y = x^{2} - 16$

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

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

$x - 4 = 0$

$y = x^{2} - 16$

Add $4$ to both sides of the equation.

$x = 4$

$y = x^{2} - 16$

$x = 4$

$y = x^{2} - 16$

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

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

$x + 5 = 0$

$y = x^{2} - 16$

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = x^{2} - 16$

$x = - 5$

$y = x^{2} - 16$

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

$x = 4, - 5$

$y = x^{2} - 16$

$x = 4, - 5$

$y = x^{2} - 16$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 16$ with $4$ .

$y = {(4)}^{2} - 16$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} - 16$ .

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

$y = 16 - 16$

$x = 4$

Subtract $16$ from $16$ .

$y = 0$

$x = 4$

$y = 0$

$x = 4$

$y = 0$

$x = 4$

$y = 0$

$x = 4$

Step 4

Replace all occurrences of $x$ with $- 5$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 16$ with $- 5$ .

$y = {(- 5)}^{2} - 16$

$x = - 5$

Simplify the right side.

Simplify ${(- 5)}^{2} - 16$ .

Raise $- 5$ to the power of $2$ .

$y = 25 - 16$

$x = - 5$

Subtract $16$ from $25$ .

$y = 9$

$x = - 5$

$y = 9$

$x = - 5$

$y = 9$

$x = - 5$

$y = 9$

$x = - 5$

Step 5

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

$(4, 0)$

$(- 5, 9)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 0), (- 5, 9)$

Equation Form:

$x = 4, y = 0$

$x = - 5, y = 9$

Step 7

Solve by Substitution x+y=4 , y=x^2-16

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