Solve the System of @WORD y=x^2-25 , x-y=-5

$y = x^{2} - 25$ , $x - y = - 5$

Step 1

Replace all occurrences of $y$ with $x^{2} - 25$ in each equation.

Replace all occurrences of $y$ in $x - y = - 5$ with $x^{2} - 25$ .

$x - (x^{2} - 25) = - 5$

$y = x^{2} - 25$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x - x^{2} + 25 = - 5$

$y = x^{2} - 25$

Multiply $- 1$ by $- 25$ .

$x - x^{2} + 25 = - 5$

$y = x^{2} - 25$

$x - x^{2} + 25 = - 5$

$y = x^{2} - 25$

$x - x^{2} + 25 = - 5$

$y = x^{2} - 25$

$x - x^{2} + 25 = - 5$

$y = x^{2} - 25$

Step 2

Solve for $x$ in $x - x^{2} + 25 = - 5$ .

Add $5$ to both sides of the equation.

$x - x^{2} + 25 + 5 = 0$

$y = x^{2} - 25$

Add $25$ and $5$ .

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

$y = x^{2} - 25$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 30$ .

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

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

$y = x^{2} - 25$

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

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

$y = x^{2} - 25$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 30 = 0$

$y = x^{2} - 25$

Rewrite $30$ as $- 1 (- 30)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 30 = 0$

$y = x^{2} - 25$

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

$- (x^{2} - x) - 1 \cdot - 30 = 0$

$y = x^{2} - 25$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 30)$ .

$- (x^{2} - x - 30) = 0$

$y = x^{2} - 25$

$- (x^{2} - x - 30) = 0$

$y = x^{2} - 25$

Factor.

Factor $x^{2} - x - 30$ 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 $- 30$ and whose sum is $- 1$ .

$- 6, 5$

$y = x^{2} - 25$

Write the factored form using these integers.

$- ((x - 6) (x + 5)) = 0$

$y = x^{2} - 25$

$- ((x - 6) (x + 5)) = 0$

$y = x^{2} - 25$

Remove unnecessary parentheses.

$- (x - 6) (x + 5) = 0$

$y = x^{2} - 25$

$- (x - 6) (x + 5) = 0$

$y = x^{2} - 25$

$- (x - 6) (x + 5) = 0$

$y = x^{2} - 25$

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

$x - 6 = 0$

$x + 5 = 0$

$y = x^{2} - 25$

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

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

$x - 6 = 0$

$y = x^{2} - 25$

Add $6$ to both sides of the equation.

$x = 6$

$y = x^{2} - 25$

$x = 6$

$y = x^{2} - 25$

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

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

$x + 5 = 0$

$y = x^{2} - 25$

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = x^{2} - 25$

$x = - 5$

$y = x^{2} - 25$

The final solution is all the values that make $- (x - 6) (x + 5) = 0$ true.

$x = 6, - 5$

$y = x^{2} - 25$

$x = 6, - 5$

$y = x^{2} - 25$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 25$ with $6$ .

$y = {(6)}^{2} - 25$

$x = 6$

Simplify the right side.

Simplify ${(6)}^{2} - 25$ .

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

$y = 36 - 25$

$x = 6$

Subtract $25$ from $36$ .

$y = 11$

$x = 6$

$y = 11$

$x = 6$

$y = 11$

$x = 6$

$y = 11$

$x = 6$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 25$ with $- 5$ .

$y = {(- 5)}^{2} - 25$

$x = - 5$

Simplify the right side.

Simplify ${(- 5)}^{2} - 25$ .

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

$y = 25 - 25$

$x = - 5$

Subtract $25$ from $25$ .

$y = 0$

$x = - 5$

$y = 0$

$x = - 5$

$y = 0$

$x = - 5$

$y = 0$

$x = - 5$

Step 5

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

$(6, 11)$

$(- 5, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, 11), (- 5, 0)$

Equation Form:

$x = 6, y = 11$

$x = - 5, y = 0$

Step 7

Solve the System of @WORD y=x^2-25 , x-y=-5

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