Solve the System of Equations y=x^2+18x+35 y=-x^2+2x+5

$y = x^{2} + 18 x + 35$ $y = - x^{2} + 2 x + 5$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} + 18 x + 35 = - x^{2} + 2 x + 5$

Step 2

Solve $x^{2} + 18 x + 35 = - x^{2} + 2 x + 5$ for $x$ .

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

Add $x^{2}$ to both sides of the equation.

$x^{2} + 18 x + 35 + x^{2} = 2 x + 5$

Subtract $2 x$ from both sides of the equation.

$x^{2} + 18 x + 35 + x^{2} - 2 x = 5$

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} + 18 x + 35 - 2 x = 5$

Subtract $2 x$ from $18 x$ .

$2 x^{2} + 16 x + 35 = 5$

Subtract $5$ from both sides of the equation.

$2 x^{2} + 16 x + 35 - 5 = 0$

Subtract $5$ from $35$ .

$2 x^{2} + 16 x + 30 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 16 x + 30$ .

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

$2 (x^{2}) + 16 x + 30 = 0$

Factor $2$ out of $16 x$ .

$2 (x^{2}) + 2 (8 x) + 30 = 0$

Factor $2$ out of $30$ .

$2 x^{2} + 2 (8 x) + 2 \cdot 15 = 0$

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

$2 (x^{2} + 8 x) + 2 \cdot 15 = 0$

Factor $2$ out of $2 (x^{2} + 8 x) + 2 \cdot 15$ .

$2 (x^{2} + 8 x + 15) = 0$

Factor.

Factor $x^{2} + 8 x + 15$ 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 $15$ and whose sum is $8$ .

$3, 5 + y = - x^{2} + 2 x + 5$

Write the factored form using these integers.

$2 ((x + 3) (x + 5)) = 0$

Remove unnecessary parentheses.

$2 (x + 3) (x + 5) = 0$

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

$x + 3 = 0$

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

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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 $2 (x + 3) (x + 5) = 0$ true.

$x = - 3, - 5$

Step 3

Evaluate $y$ when $x = - 3$ .

Substitute $- 3$ for $x$ .

$y = - {(- 3)}^{2} + 2 (- 3) + 5$

Simplify $- {(- 3)}^{2} + 2 (- 3) + 5$ .

Simplify each term.

Raise $- 3$ to the power of $2$ .

$y = - 1 \cdot 9 + 2 (- 3) + 5$

Multiply $- 1$ by $9$ .

$y = - 9 + 2 (- 3) + 5$

Multiply $2$ by $- 3$ .

$y = - 9 - 6 + 5$

Simplify by adding and subtracting.

Subtract $6$ from $- 9$ .

$y = - 15 + 5$

Add $- 15$ and $5$ .

$y = - 10$

Step 4

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

Substitute $- 5$ for $x$ .

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

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

Simplify each term.

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

$y = - 1 \cdot 25 + 2 (- 5) + 5$

Multiply $- 1$ by $25$ .

$y = - 25 + 2 (- 5) + 5$

Multiply $2$ by $- 5$ .

$y = - 25 - 10 + 5$

Simplify by adding and subtracting.

Subtract $10$ from $- 25$ .

$y = - 35 + 5$

Add $- 35$ and $5$ .

$y = - 30$

Step 5

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

$(- 3, - 10)$

$(- 5, - 30)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 3, - 10), (- 5, - 30)$

Equation Form:

$x = - 3, y = - 10$

$x = - 5, y = - 30$

Step 7

Solve the System of Equations y=x^2+18x+35 y=-x^2+2x+5

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