Solve the System of Equations y-15=-x^2+4x x+y=1

$y - 15 = - x^{2} + 4 x$ $x + y = 1$

Step 1

Add $15$ to both sides of the equation.

$y = - x^{2} + 4 x + 15$

$x + y = 1$

Step 2

Replace all occurrences of $y$ with $- x^{2} + 4 x + 15$ in each equation.

Replace all occurrences of $y$ in $x + y = 1$ with $- x^{2} + 4 x + 15$ .

$x - x^{2} + 4 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

Simplify the left side.

Simplify $x - x^{2} + 4 x + 15$ .

Remove parentheses.

$x - x^{2} + 4 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

Add $x$ and $4 x$ .

$- x^{2} + 5 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

$- x^{2} + 5 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

$- x^{2} + 5 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

$- x^{2} + 5 x + 15 = 1$

$y = - x^{2} + 4 x + 15$

Step 3

Solve for $x$ in $- x^{2} + 5 x + 15 = 1$ .

Subtract $1$ from both sides of the equation.

$- x^{2} + 5 x + 15 - 1 = 0$

$y = - x^{2} + 4 x + 15$

Subtract $1$ from $15$ .

$- x^{2} + 5 x + 14 = 0$

$y = - x^{2} + 4 x + 15$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 5 x + 14$ .

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

$- (x^{2}) + 5 x + 14 = 0$

$y = - x^{2} + 4 x + 15$

Factor $- 1$ out of $5 x$ .

$- (x^{2}) - (- 5 x) + 14 = 0$

$y = - x^{2} + 4 x + 15$

Rewrite $14$ as $- 1 (- 14)$ .

$- (x^{2}) - (- 5 x) - 1 \cdot - 14 = 0$

$y = - x^{2} + 4 x + 15$

Factor $- 1$ out of $- (x^{2}) - (- 5 x)$ .

$- (x^{2} - 5 x) - 1 \cdot - 14 = 0$

$y = - x^{2} + 4 x + 15$

Factor $- 1$ out of $- (x^{2} - 5 x) - 1 (- 14)$ .

$- (x^{2} - 5 x - 14) = 0$

$y = - x^{2} + 4 x + 15$

$- (x^{2} - 5 x - 14) = 0$

$y = - x^{2} + 4 x + 15$

Factor.

Factor $x^{2} - 5 x - 14$ 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 $- 14$ and whose sum is $- 5$ .

$- 7, 2$

$y = - x^{2} + 4 x + 15$

Write the factored form using these integers.

$- ((x - 7) (x + 2)) = 0$

$y = - x^{2} + 4 x + 15$

$- ((x - 7) (x + 2)) = 0$

$y = - x^{2} + 4 x + 15$

Remove unnecessary parentheses.

$- (x - 7) (x + 2) = 0$

$y = - x^{2} + 4 x + 15$

$- (x - 7) (x + 2) = 0$

$y = - x^{2} + 4 x + 15$

$- (x - 7) (x + 2) = 0$

$y = - x^{2} + 4 x + 15$

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

$x - 7 = 0$

$x + 2 = 0$

$y = - x^{2} + 4 x + 15$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

$y = - x^{2} + 4 x + 15$

Add $7$ to both sides of the equation.

$x = 7$

$y = - x^{2} + 4 x + 15$

$x = 7$

$y = - x^{2} + 4 x + 15$

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

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

$x + 2 = 0$

$y = - x^{2} + 4 x + 15$

Subtract $2$ from both sides of the equation.

$x = - 2$

$y = - x^{2} + 4 x + 15$

$x = - 2$

$y = - x^{2} + 4 x + 15$

The final solution is all the values that make $- (x - 7) (x + 2) = 0$ true.

$x = 7, - 2$

$y = - x^{2} + 4 x + 15$

$x = 7, - 2$

$y = - x^{2} + 4 x + 15$

Step 4

Replace all occurrences of $x$ with $7$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} + 4 x + 15$ with $7$ .

$y = - {(7)}^{2} + 4 (7) + 15$

$x = 7$

Simplify the right side.

Simplify $- {(7)}^{2} + 4 (7) + 15$ .

Simplify each term.

Raise $7$ to the power of $2$ .

$y = - 1 \cdot 49 + 4 (7) + 15$

$x = 7$

Multiply $- 1$ by $49$ .

$y = - 49 + 4 (7) + 15$

$x = 7$

Multiply $4$ by $7$ .

$y = - 49 + 28 + 15$

$x = 7$

$y = - 49 + 28 + 15$

$x = 7$

Simplify by adding numbers.

Add $- 49$ and $28$ .

$y = - 21 + 15$

$x = 7$

Add $- 21$ and $15$ .

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

Step 5

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

Replace all occurrences of $x$ in $y = - x^{2} + 4 x + 15$ with $- 2$ .

$y = - {(- 2)}^{2} + 4 (- 2) + 15$

$x = - 2$

Simplify the right side.

Simplify $- {(- 2)}^{2} + 4 (- 2) + 15$ .

Simplify each term.

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

$y = - 1 \cdot 4 + 4 (- 2) + 15$

$x = - 2$

Multiply $- 1$ by $4$ .

$y = - 4 + 4 (- 2) + 15$

$x = - 2$

Multiply $4$ by $- 2$ .

$y = - 4 - 8 + 15$

$x = - 2$

$y = - 4 - 8 + 15$

$x = - 2$

Simplify by adding and subtracting.

Subtract $8$ from $- 4$ .

$y = - 12 + 15$

$x = - 2$

Add $- 12$ and $15$ .

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

Step 6

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

$(7, - 6)$

$(- 2, 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(7, - 6), (- 2, 3)$

Equation Form:

$x = 7, y = - 6$

$x = - 2, y = 3$

Step 8

Solve the System of Equations y-15=-x^2+4x x+y=1

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