Solve the System of Equations y=x^2+6x -x+y=24

$y = x^{2} + 6 x$ $- x + y = 24$

Replace all occurrences of $y$ with $x^{2} + 6 x$ in each equation.

Replace all occurrences of $y$ in $- x + y = 24$ with $x^{2} + 6 x$ .

$- x + x^{2} + 6 x = 24$

$y = x^{2} + 6 x$

Simplify the left side.

Simplify $- x + x^{2} + 6 x$ .

Remove parentheses.

$- x + x^{2} + 6 x = 24$

$y = x^{2} + 6 x$

Add $- x$ and $6 x$ .

$x^{2} + 5 x = 24$

$y = x^{2} + 6 x$

$x^{2} + 5 x = 24$

$y = x^{2} + 6 x$

$x^{2} + 5 x = 24$

$y = x^{2} + 6 x$

$x^{2} + 5 x = 24$

$y = x^{2} + 6 x$

Solve for $x$ in $x^{2} + 5 x = 24$ .

Subtract $24$ from both sides of the equation.

$x^{2} + 5 x - 24 = 0$

$y = x^{2} + 6 x$

Factor $x^{2} + 5 x - 24$ 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 $5$ .

$- 3, 8$

$y = x^{2} + 6 x$

Write the factored form using these integers.

$(x - 3) (x + 8) = 0$

$y = x^{2} + 6 x$

$(x - 3) (x + 8) = 0$

$y = x^{2} + 6 x$

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

$x - 3 = 0$

$x + 8 = 0$

$y = x^{2} + 6 x$

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

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

$x - 3 = 0$

$y = x^{2} + 6 x$

Add $3$ to both sides of the equation.

$x = 3$

$y = x^{2} + 6 x$

$x = 3$

$y = x^{2} + 6 x$

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

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

$x + 8 = 0$

$y = x^{2} + 6 x$

Subtract $8$ from both sides of the equation.

$x = - 8$

$y = x^{2} + 6 x$

$x = - 8$

$y = x^{2} + 6 x$

The final solution is all the values that make $(x - 3) (x + 8) = 0$ true.

$x = 3, - 8$

$y = x^{2} + 6 x$

$x = 3, - 8$

$y = x^{2} + 6 x$

Replace all occurrences of $x$ with $3$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} + 6 x$ with $3$ .

$y = {(3)}^{2} + 6 (3)$

$x = 3$

Simplify the right side.

Simplify ${(3)}^{2} + 6 (3)$ .

Simplify each term.

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

$y = 9 + 6 (3)$

$x = 3$

Multiply $6$ by $3$ .

$y = 9 + 18$

$x = 3$

$y = 9 + 18$

$x = 3$

Add $9$ and $18$ .

$y = 27$

$x = 3$

$y = 27$

$x = 3$

$y = 27$

$x = 3$

$y = 27$

$x = 3$

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

Replace all occurrences of $x$ in $y = x^{2} + 6 x$ with $- 8$ .

$y = {(- 8)}^{2} + 6 (- 8)$

$x = - 8$

Simplify the right side.

Simplify ${(- 8)}^{2} + 6 (- 8)$ .

Simplify each term.

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

$y = 64 + 6 (- 8)$

$x = - 8$

Multiply $6$ by $- 8$ .

$y = 64 - 48$

$x = - 8$

$y = 64 - 48$

$x = - 8$

Subtract $48$ from $64$ .

$y = 16$

$x = - 8$

$y = 16$

$x = - 8$

$y = 16$

$x = - 8$

$y = 16$

$x = - 8$

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

$(3, 27)$

$(- 8, 16)$

The result can be shown in multiple forms.

Point Form:

$(3, 27), (- 8, 16)$

Equation Form:

$x = 3, y = 27$

$x = - 8, y = 16$

Solve the System of Equations y=x^2+6x -x+y=24

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