Solve the System of @WORD y=x^2-49 , x-y=-7

$y = x^{2} - 49$ , $x - y = - 7$

Step 1

Replace all occurrences of $y$ with $x^{2} - 49$ in each equation.

Replace all occurrences of $y$ in $x - y = - 7$ with $x^{2} - 49$ .

$x - (x^{2} - 49) = - 7$

$y = x^{2} - 49$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x - x^{2} + 49 = - 7$

$y = x^{2} - 49$

Multiply $- 1$ by $- 49$ .

$x - x^{2} + 49 = - 7$

$y = x^{2} - 49$

$x - x^{2} + 49 = - 7$

$y = x^{2} - 49$

$x - x^{2} + 49 = - 7$

$y = x^{2} - 49$

$x - x^{2} + 49 = - 7$

$y = x^{2} - 49$

Step 2

Solve for $x$ in $x - x^{2} + 49 = - 7$ .

Add $7$ to both sides of the equation.

$x - x^{2} + 49 + 7 = 0$

$y = x^{2} - 49$

Add $49$ and $7$ .

$x - x^{2} + 56 = 0$

$y = x^{2} - 49$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 56$ .

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

$- x^{2} + x + 56 = 0$

$y = x^{2} - 49$

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

$- (x^{2}) + x + 56 = 0$

$y = x^{2} - 49$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 56 = 0$

$y = x^{2} - 49$

Rewrite $56$ as $- 1 (- 56)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 56 = 0$

$y = x^{2} - 49$

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

$- (x^{2} - x) - 1 \cdot - 56 = 0$

$y = x^{2} - 49$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 56)$ .

$- (x^{2} - x - 56) = 0$

$y = x^{2} - 49$

$- (x^{2} - x - 56) = 0$

$y = x^{2} - 49$

Factor.

Factor $x^{2} - x - 56$ 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 $- 56$ and whose sum is $- 1$ .

$- 8, 7$

$y = x^{2} - 49$

Write the factored form using these integers.

$- ((x - 8) (x + 7)) = 0$

$y = x^{2} - 49$

$- ((x - 8) (x + 7)) = 0$

$y = x^{2} - 49$

Remove unnecessary parentheses.

$- (x - 8) (x + 7) = 0$

$y = x^{2} - 49$

$- (x - 8) (x + 7) = 0$

$y = x^{2} - 49$

$- (x - 8) (x + 7) = 0$

$y = x^{2} - 49$

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

$x - 8 = 0$

$x + 7 = 0$

$y = x^{2} - 49$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

$y = x^{2} - 49$

Add $8$ to both sides of the equation.

$x = 8$

$y = x^{2} - 49$

$x = 8$

$y = x^{2} - 49$

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

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

$x + 7 = 0$

$y = x^{2} - 49$

Subtract $7$ from both sides of the equation.

$x = - 7$

$y = x^{2} - 49$

$x = - 7$

$y = x^{2} - 49$

The final solution is all the values that make $- (x - 8) (x + 7) = 0$ true.

$x = 8, - 7$

$y = x^{2} - 49$

$x = 8, - 7$

$y = x^{2} - 49$

Step 3

Replace all occurrences of $x$ with $8$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 49$ with $8$ .

$y = {(8)}^{2} - 49$

$x = 8$

Simplify the right side.

Simplify ${(8)}^{2} - 49$ .

Raise $8$ to the power of $2$ .

$y = 64 - 49$

$x = 8$

Subtract $49$ from $64$ .

$y = 15$

$x = 8$

$y = 15$

$x = 8$

$y = 15$

$x = 8$

$y = 15$

$x = 8$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 49$ with $- 7$ .

$y = {(- 7)}^{2} - 49$

$x = - 7$

Simplify the right side.

Simplify ${(- 7)}^{2} - 49$ .

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

$y = 49 - 49$

$x = - 7$

Subtract $49$ from $49$ .

$y = 0$

$x = - 7$

$y = 0$

$x = - 7$

$y = 0$

$x = - 7$

$y = 0$

$x = - 7$

Step 5

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

$(8, 15)$

$(- 7, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(8, 15), (- 7, 0)$

Equation Form:

$x = 8, y = 15$

$x = - 7, y = 0$

Step 7

Solve the System of @WORD y=x^2-49 , x-y=-7

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