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

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

Step 1

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

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

$y + 4 x = 16$

Step 2

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

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

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

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

Simplify the left side.

Add $4 x$ and $4 x$ .

$- x^{2} + 8 x = 16$

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

$- x^{2} + 8 x = 16$

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

$- x^{2} + 8 x = 16$

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

Step 3

Solve for $x$ in $- x^{2} + 8 x = 16$ .

Subtract $16$ from both sides of the equation.

$- x^{2} + 8 x - 16 = 0$

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 8 x - 16$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 8 x - 16 = 0$

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

Factor $- 1$ out of $8 x$ .

$- (x^{2}) - (- 8 x) - 16 = 0$

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

Rewrite $- 16$ as $- 1 (16)$ .

$- (x^{2}) - (- 8 x) - 1 \cdot 16 = 0$

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

Factor $- 1$ out of $- (x^{2}) - (- 8 x)$ .

$- (x^{2} - 8 x) - 1 \cdot 16 = 0$

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

Factor $- 1$ out of $- (x^{2} - 8 x) - 1 (16)$ .

$- (x^{2} - 8 x + 16) = 0$

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

$- (x^{2} - 8 x + 16) = 0$

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

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

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

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 4 + 4^{2}) = 0$

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$- {(x - 4)}^{2} = 0$

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

$- {(x - 4)}^{2} = 0$

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

$- {(x - 4)}^{2} = 0$

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

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 4)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 4)}^{2}}{- 1} = \frac{0}{- 1}$

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

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 4)}^{2}}{1} = \frac{0}{- 1}$

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

Divide ${(x - 4)}^{2}$ by $1$ .

${(x - 4)}^{2} = \frac{0}{- 1}$

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

${(x - 4)}^{2} = \frac{0}{- 1}$

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

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 4)}^{2} = 0$

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

${(x - 4)}^{2} = 0$

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

${(x - 4)}^{2} = 0$

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

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

$x - 4 = 0$

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

Add $4$ to both sides of the equation.

$x = 4$

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

$x = 4$

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

Step 4

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

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

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

$x = 4$

Simplify the right side.

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

Simplify each term.

Multiply $4$ by $4$ .

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

$x = 4$

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

$y = 16 - 1 \cdot 16$

$x = 4$

Multiply $- 1$ by $16$ .

$y = 16 - 16$

$x = 4$

$y = 16 - 16$

$x = 4$

Subtract $16$ from $16$ .

$y = 0$

$x = 4$

$y = 0$

$x = 4$

$y = 0$

$x = 4$

$y = 0$

$x = 4$

Step 5

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

$(4, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 0)$

Equation Form:

$x = 4, y = 0$

Step 7

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

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