Solve the System of Equations y=-x^2+13x-40 x+y=5

$y = - x^{2} + 13 x - 40$ $x + y = 5$

Step 1

Replace all occurrences of $y$ with $- x^{2} + 13 x - 40$ in each equation.

Replace all occurrences of $y$ in $x + y = 5$ with $- x^{2} + 13 x - 40$ .

$x - x^{2} + 13 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

Simplify the left side.

Simplify $x - x^{2} + 13 x - 40$ .

Remove parentheses.

$x - x^{2} + 13 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

Add $x$ and $13 x$ .

$- x^{2} + 14 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

$- x^{2} + 14 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

$- x^{2} + 14 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

$- x^{2} + 14 x - 40 = 5$

$y = - x^{2} + 13 x - 40$

Step 2

Solve for $x$ in $- x^{2} + 14 x - 40 = 5$ .

Subtract $5$ from both sides of the equation.

$- x^{2} + 14 x - 40 - 5 = 0$

$y = - x^{2} + 13 x - 40$

Subtract $5$ from $- 40$ .

$- x^{2} + 14 x - 45 = 0$

$y = - x^{2} + 13 x - 40$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 14 x - 45$ .

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

$- (x^{2}) + 14 x - 45 = 0$

$y = - x^{2} + 13 x - 40$

Factor $- 1$ out of $14 x$ .

$- (x^{2}) - (- 14 x) - 45 = 0$

$y = - x^{2} + 13 x - 40$

Rewrite $- 45$ as $- 1 (45)$ .

$- (x^{2}) - (- 14 x) - 1 \cdot 45 = 0$

$y = - x^{2} + 13 x - 40$

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

$- (x^{2} - 14 x) - 1 \cdot 45 = 0$

$y = - x^{2} + 13 x - 40$

Factor $- 1$ out of $- (x^{2} - 14 x) - 1 (45)$ .

$- (x^{2} - 14 x + 45) = 0$

$y = - x^{2} + 13 x - 40$

$- (x^{2} - 14 x + 45) = 0$

$y = - x^{2} + 13 x - 40$

Factor.

Factor $x^{2} - 14 x + 45$ 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 $45$ and whose sum is $- 14$ .

$- 9, - 5$

$y = - x^{2} + 13 x - 40$

Write the factored form using these integers.

$- ((x - 9) (x - 5)) = 0$

$y = - x^{2} + 13 x - 40$

$- ((x - 9) (x - 5)) = 0$

$y = - x^{2} + 13 x - 40$

Remove unnecessary parentheses.

$- (x - 9) (x - 5) = 0$

$y = - x^{2} + 13 x - 40$

$- (x - 9) (x - 5) = 0$

$y = - x^{2} + 13 x - 40$

$- (x - 9) (x - 5) = 0$

$y = - x^{2} + 13 x - 40$

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

$x - 9 = 0$

$x - 5 = 0$

$y = - x^{2} + 13 x - 40$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

$y = - x^{2} + 13 x - 40$

Add $9$ to both sides of the equation.

$x = 9$

$y = - x^{2} + 13 x - 40$

$x = 9$

$y = - x^{2} + 13 x - 40$

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

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

$x - 5 = 0$

$y = - x^{2} + 13 x - 40$

Add $5$ to both sides of the equation.

$x = 5$

$y = - x^{2} + 13 x - 40$

$x = 5$

$y = - x^{2} + 13 x - 40$

The final solution is all the values that make $- (x - 9) (x - 5) = 0$ true.

$x = 9, 5$

$y = - x^{2} + 13 x - 40$

$x = 9, 5$

$y = - x^{2} + 13 x - 40$

Step 3

Replace all occurrences of $x$ with $9$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} + 13 x - 40$ with $9$ .

$y = - {(9)}^{2} + 13 (9) - 40$

$x = 9$

Simplify the right side.

Simplify $- {(9)}^{2} + 13 (9) - 40$ .

Simplify each term.

Raise $9$ to the power of $2$ .

$y = - 1 \cdot 81 + 13 (9) - 40$

$x = 9$

Multiply $- 1$ by $81$ .

$y = - 81 + 13 (9) - 40$

$x = 9$

Multiply $13$ by $9$ .

$y = - 81 + 117 - 40$

$x = 9$

$y = - 81 + 117 - 40$

$x = 9$

Simplify by adding and subtracting.

Add $- 81$ and $117$ .

$y = 36 - 40$

$x = 9$

Subtract $40$ from $36$ .

$y = - 4$

$x = 9$

$y = - 4$

$x = 9$

$y = - 4$

$x = 9$

$y = - 4$

$x = 9$

$y = - 4$

$x = 9$

Step 4

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

Replace all occurrences of $x$ in $y = - x^{2} + 13 x - 40$ with $5$ .

$y = - {(5)}^{2} + 13 (5) - 40$

$x = 5$

Simplify the right side.

Simplify $- {(5)}^{2} + 13 (5) - 40$ .

Simplify each term.

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

$y = - 1 \cdot 25 + 13 (5) - 40$

$x = 5$

Multiply $- 1$ by $25$ .

$y = - 25 + 13 (5) - 40$

$x = 5$

Multiply $13$ by $5$ .

$y = - 25 + 65 - 40$

$x = 5$

$y = - 25 + 65 - 40$

$x = 5$

Simplify by adding and subtracting.

Add $- 25$ and $65$ .

$y = 40 - 40$

$x = 5$

Subtract $40$ from $40$ .

$y = 0$

$x = 5$

$y = 0$

$x = 5$

$y = 0$

$x = 5$

$y = 0$

$x = 5$

$y = 0$

$x = 5$

Step 5

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

$(9, - 4)$

$(5, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 9, y = - 4$

$x = 5, y = 0$

Step 7

Solve the System of Equations y=-x^2+13x-40 x+y=5

Related Questions

Support LCMGCF.com by disabling your adblocker.