Solve the System of @WORD y=x^2-36 , x-y=-6

$y = x^{2} - 36$ , $x - y = - 6$

Step 1

Replace all occurrences of $y$ with $x^{2} - 36$ in each equation.

Replace all occurrences of $y$ in $x - y = - 6$ with $x^{2} - 36$ .

$x - (x^{2} - 36) = - 6$

$y = x^{2} - 36$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x - x^{2} + 36 = - 6$

$y = x^{2} - 36$

Multiply $- 1$ by $- 36$ .

$x - x^{2} + 36 = - 6$

$y = x^{2} - 36$

$x - x^{2} + 36 = - 6$

$y = x^{2} - 36$

$x - x^{2} + 36 = - 6$

$y = x^{2} - 36$

$x - x^{2} + 36 = - 6$

$y = x^{2} - 36$

Step 2

Solve for $x$ in $x - x^{2} + 36 = - 6$ .

Add $6$ to both sides of the equation.

$x - x^{2} + 36 + 6 = 0$

$y = x^{2} - 36$

Add $36$ and $6$ .

$x - x^{2} + 42 = 0$

$y = x^{2} - 36$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 42$ .

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 42 = 0$

$y = x^{2} - 36$

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

$- (x^{2}) + x + 42 = 0$

$y = x^{2} - 36$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 42 = 0$

$y = x^{2} - 36$

Rewrite $42$ as $- 1 (- 42)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 42 = 0$

$y = x^{2} - 36$

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

$- (x^{2} - x) - 1 \cdot - 42 = 0$

$y = x^{2} - 36$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 42)$ .

$- (x^{2} - x - 42) = 0$

$y = x^{2} - 36$

$- (x^{2} - x - 42) = 0$

$y = x^{2} - 36$

Factor.

Factor $x^{2} - x - 42$ 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 $- 42$ and whose sum is $- 1$ .

$- 7, 6$

$y = x^{2} - 36$

Write the factored form using these integers.

$- ((x - 7) (x + 6)) = 0$

$y = x^{2} - 36$

$- ((x - 7) (x + 6)) = 0$

$y = x^{2} - 36$

Remove unnecessary parentheses.

$- (x - 7) (x + 6) = 0$

$y = x^{2} - 36$

$- (x - 7) (x + 6) = 0$

$y = x^{2} - 36$

$- (x - 7) (x + 6) = 0$

$y = x^{2} - 36$

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

$x - 7 = 0$

$x + 6 = 0$

$y = x^{2} - 36$

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

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

$x - 7 = 0$

$y = x^{2} - 36$

Add $7$ to both sides of the equation.

$x = 7$

$y = x^{2} - 36$

$x = 7$

$y = x^{2} - 36$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

$y = x^{2} - 36$

Subtract $6$ from both sides of the equation.

$x = - 6$

$y = x^{2} - 36$

$x = - 6$

$y = x^{2} - 36$

The final solution is all the values that make $- (x - 7) (x + 6) = 0$ true.

$x = 7, - 6$

$y = x^{2} - 36$

$x = 7, - 6$

$y = x^{2} - 36$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 36$ with $7$ .

$y = {(7)}^{2} - 36$

$x = 7$

Simplify the right side.

Simplify ${(7)}^{2} - 36$ .

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

$y = 49 - 36$

$x = 7$

Subtract $36$ from $49$ .

$y = 13$

$x = 7$

$y = 13$

$x = 7$

$y = 13$

$x = 7$

$y = 13$

$x = 7$

Step 4

Replace all occurrences of $x$ with $- 6$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 36$ with $- 6$ .

$y = {(- 6)}^{2} - 36$

$x = - 6$

Simplify the right side.

Simplify ${(- 6)}^{2} - 36$ .

Raise $- 6$ to the power of $2$ .

$y = 36 - 36$

$x = - 6$

Subtract $36$ from $36$ .

$y = 0$

$x = - 6$

$y = 0$

$x = - 6$

$y = 0$

$x = - 6$

$y = 0$

$x = - 6$

Step 5

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

$(7, 13)$

$(- 6, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(7, 13), (- 6, 0)$

Equation Form:

$x = 7, y = 13$

$x = - 6, y = 0$

Step 7

Solve the System of @WORD y=x^2-36 , x-y=-6

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