Solve the System of Equations y=x^2-5x x+y=21

$y = x^{2} - 5 x$ $x + y = 21$

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

Replace all occurrences of $y$ in $x + y = 21$ with $x^{2} - 5 x$ .

$x + x^{2} - 5 x = 21$

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

Simplify the left side.

Simplify $x + x^{2} - 5 x$ .

Remove parentheses.

$x + x^{2} - 5 x = 21$

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

Subtract $5 x$ from $x$ .

$x^{2} - 4 x = 21$

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

$x^{2} - 4 x = 21$

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

$x^{2} - 4 x = 21$

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

$x^{2} - 4 x = 21$

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

Solve for $x$ in $x^{2} - 4 x = 21$ .

Subtract $21$ from both sides of the equation.

$x^{2} - 4 x - 21 = 0$

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

Factor $x^{2} - 4 x - 21$ 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 $- 21$ and whose sum is $- 4$ .

$- 7, 3$

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

Write the factored form using these integers.

$(x - 7) (x + 3) = 0$

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

$(x - 7) (x + 3) = 0$

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

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

$x - 7 = 0$

$x + 3 = 0$

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

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

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

$x - 7 = 0$

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

Add $7$ to both sides of the equation.

$x = 7$

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

$x = 7$

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

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

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

$x + 3 = 0$

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

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = - 3$

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

The final solution is all the values that make $(x - 7) (x + 3) = 0$ true.

$x = 7, - 3$

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

$x = 7, - 3$

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

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

Replace all occurrences of $x$ in $y = x^{2} - 5 x$ with $7$ .

$y = {(7)}^{2} - 5 \cdot 7$

$x = 7$

Simplify the right side.

Simplify ${(7)}^{2} - 5 \cdot 7$ .

Simplify each term.

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

$y = 49 - 5 \cdot 7$

$x = 7$

Multiply $- 5$ by $7$ .

$y = 49 - 35$

$x = 7$

$y = 49 - 35$

$x = 7$

Subtract $35$ from $49$ .

$y = 14$

$x = 7$

$y = 14$

$x = 7$

$y = 14$

$x = 7$

$y = 14$

$x = 7$

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

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

$y = {(- 3)}^{2} - 5 \cdot - 3$

$x = - 3$

Simplify the right side.

Simplify ${(- 3)}^{2} - 5 \cdot - 3$ .

Simplify each term.

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

$y = 9 - 5 \cdot - 3$

$x = - 3$

Multiply $- 5$ by $- 3$ .

$y = 9 + 15$

$x = - 3$

$y = 9 + 15$

$x = - 3$

Add $9$ and $15$ .

$y = 24$

$x = - 3$

$y = 24$

$x = - 3$

$y = 24$

$x = - 3$

$y = 24$

$x = - 3$

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

$(7, 14)$

$(- 3, 24)$

The result can be shown in multiple forms.

Point Form:

$(7, 14), (- 3, 24)$

Equation Form:

$x = 7, y = 14$

$x = - 3, y = 24$

Solve the System of Equations y=x^2-5x x+y=21

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