Solve the System of Equations y=x^2+5x-25 1=3x-y

$y = x^{2} + 5 x - 25$ $1 = 3 x - y$

Step 1

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

Replace all occurrences of $y$ in $1 = 3 x - y$ with $x^{2} + 5 x - 25$ .

$1 = 3 x - (x^{2} + 5 x - 25)$

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

Simplify the right side.

Simplify $3 x - (x^{2} + 5 x - 25)$ .

Simplify each term.

Apply the distributive property.

$1 = 3 x - x^{2} - (5 x) + 25$

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

Simplify.

Multiply $5$ by $- 1$ .

$1 = 3 x - x^{2} - 5 x + 25$

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

Multiply $- 1$ by $- 25$ .

$1 = 3 x - x^{2} - 5 x + 25$

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

$1 = 3 x - x^{2} - 5 x + 25$

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

$1 = 3 x - x^{2} - 5 x + 25$

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

Subtract $5 x$ from $3 x$ .

$1 = - x^{2} - 2 x + 25$

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

$1 = - x^{2} - 2 x + 25$

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

$1 = - x^{2} - 2 x + 25$

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

$1 = - x^{2} - 2 x + 25$

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

Step 2

Solve for $x$ in $1 = - x^{2} - 2 x + 25$ .

Rewrite the equation as $- x^{2} - 2 x + 25 = 1$ .

$- x^{2} - 2 x + 25 = 1$

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

Subtract $1$ from both sides of the equation.

$- x^{2} - 2 x + 25 - 1 = 0$

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

Subtract $1$ from $25$ .

$- x^{2} - 2 x + 24 = 0$

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 2 x + 24$ .

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

$- (x^{2}) - 2 x + 24 = 0$

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

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 24 = 0$

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

Rewrite $24$ as $- 1 (- 24)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 24 = 0$

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

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

$- (x^{2} + 2 x) - 1 \cdot - 24 = 0$

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

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 24)$ .

$- (x^{2} + 2 x - 24) = 0$

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

$- (x^{2} + 2 x - 24) = 0$

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

Factor.

Factor $x^{2} + 2 x - 24$ 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

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

Write the factored form using these integers.

$- ((x - 4) (x + 6)) = 0$

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

$- ((x - 4) (x + 6)) = 0$

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

Remove unnecessary parentheses.

$- (x - 4) (x + 6) = 0$

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

$- (x - 4) (x + 6) = 0$

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

$- (x - 4) (x + 6) = 0$

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

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

$x - 4 = 0$

$x + 6 = 0$

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

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

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

Add $4$ to both sides of the equation.

$x = 4$

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

$x = 4$

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

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

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

$x + 6 = 0$

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

Subtract $6$ from both sides of the equation.

$x = - 6$

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

$x = - 6$

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

The final solution is all the values that make $- (x - 4) (x + 6) = 0$ true.

$x = 4, - 6$

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

$x = 4, - 6$

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

Step 3

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} + 5 x - 25$ with $4$ .

$y = {(4)}^{2} + 5 (4) - 25$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} + 5 (4) - 25$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$y = 16 + 5 (4) - 25$

$x = 4$

Multiply $5$ by $4$ .

$y = 16 + 20 - 25$

$x = 4$

$y = 16 + 20 - 25$

$x = 4$

Simplify by adding and subtracting.

Add $16$ and $20$ .

$y = 36 - 25$

$x = 4$

Subtract $25$ from $36$ .

$y = 11$

$x = 4$

$y = 11$

$x = 4$

$y = 11$

$x = 4$

$y = 11$

$x = 4$

$y = 11$

$x = 4$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} + 5 x - 25$ with $- 6$ .

$y = {(- 6)}^{2} + 5 (- 6) - 25$

$x = - 6$

Simplify the right side.

Simplify ${(- 6)}^{2} + 5 (- 6) - 25$ .

Simplify each term.

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

$y = 36 + 5 (- 6) - 25$

$x = - 6$

Multiply $5$ by $- 6$ .

$y = 36 - 30 - 25$

$x = - 6$

$y = 36 - 30 - 25$

$x = - 6$

Simplify by subtracting numbers.

Subtract $30$ from $36$ .

$y = 6 - 25$

$x = - 6$

Subtract $25$ from $6$ .

$y = - 19$

$x = - 6$

$y = - 19$

$x = - 6$

$y = - 19$

$x = - 6$

$y = - 19$

$x = - 6$

$y = - 19$

$x = - 6$

Step 5

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

$(4, 11)$

$(- 6, - 19)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 11), (- 6, - 19)$

Equation Form:

$x = 4, y = 11$

$x = - 6, y = - 19$

Step 7

Solve the System of Equations y=x^2+5x-25 1=3x-y

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