Algebra Examples

$y = x^{2} + x - 27$ $7 = 2 x - y$

Step 1

Replace all occurrences of $y$ with $x^{2} + x - 27$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $7 = 2 x - y$ with $x^{2} + x - 27$ .

$7 = 2 x - (x^{2} + x - 27)$

$y = x^{2} + x - 27$

Step 1.2

Simplify the right side.

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Step 1.2.1

Simplify $2 x - (x^{2} + x - 27)$ .

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Step 1.2.1.1

Simplify each term.

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Step 1.2.1.1.1

Apply the distributive property.

$7 = 2 x - x^{2} - x + 27$

$y = x^{2} + x - 27$

Step 1.2.1.1.2

Multiply $- 1$ by $- 27$ .

$7 = 2 x - x^{2} - x + 27$

$y = x^{2} + x - 27$

$7 = 2 x - x^{2} - x + 27$

$y = x^{2} + x - 27$

Step 1.2.1.2

Subtract $x$ from $2 x$ .

$7 = - x^{2} + x + 27$

$y = x^{2} + x - 27$

$7 = - x^{2} + x + 27$

$y = x^{2} + x - 27$

$7 = - x^{2} + x + 27$

$y = x^{2} + x - 27$

$7 = - x^{2} + x + 27$

$y = x^{2} + x - 27$

Step 2

Solve for $x$ in $7 = - x^{2} + x + 27$ .

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Step 2.1

Rewrite the equation as $- x^{2} + x + 27 = 7$ .

$- x^{2} + x + 27 = 7$

$y = x^{2} + x - 27$

Step 2.2

Subtract $7$ from both sides of the equation.

$- x^{2} + x + 27 - 7 = 0$

$y = x^{2} + x - 27$

Step 2.3

Subtract $7$ from $27$ .

$- x^{2} + x + 20 = 0$

$y = x^{2} + x - 27$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $- 1$ out of $- x^{2} + x + 20$ .

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Step 2.4.1.1

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

$- (x^{2}) + x + 20 = 0$

$y = x^{2} + x - 27$

Step 2.4.1.2

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 20 = 0$

$y = x^{2} + x - 27$

Step 2.4.1.3

Rewrite $20$ as $- 1 (- 20)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 20 = 0$

$y = x^{2} + x - 27$

Step 2.4.1.4

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

$- (x^{2} - x) - 1 \cdot - 20 = 0$

$y = x^{2} + x - 27$

Step 2.4.1.5

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 20)$ .

$- (x^{2} - x - 20) = 0$

$y = x^{2} + x - 27$

$- (x^{2} - x - 20) = 0$

$y = x^{2} + x - 27$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} - x - 20$ using the AC method.

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Step 2.4.2.1.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

$y = x^{2} + x - 27$

Step 2.4.2.1.2

Write the factored form using these integers.

$- ((x - 5) (x + 4)) = 0$

$y = x^{2} + x - 27$

$- ((x - 5) (x + 4)) = 0$

$y = x^{2} + x - 27$

Step 2.4.2.2

Remove unnecessary parentheses.

$- (x - 5) (x + 4) = 0$

$y = x^{2} + x - 27$

$- (x - 5) (x + 4) = 0$

$y = x^{2} + x - 27$

$- (x - 5) (x + 4) = 0$

$y = x^{2} + x - 27$

Step 2.5

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

$x - 5 = 0$

$x + 4 = 0$

$y = x^{2} + x - 27$

Step 2.6

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

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Step 2.6.1

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

$x - 5 = 0$

$y = x^{2} + x - 27$

Step 2.6.2

Add $5$ to both sides of the equation.

$x = 5$

$y = x^{2} + x - 27$

$x = 5$

$y = x^{2} + x - 27$

Step 2.7

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

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Step 2.7.1

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

$x + 4 = 0$

$y = x^{2} + x - 27$

Step 2.7.2

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = x^{2} + x - 27$

$x = - 4$

$y = x^{2} + x - 27$

Step 2.8

The final solution is all the values that make $- (x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

$y = x^{2} + x - 27$

$x = 5, - 4$

$y = x^{2} + x - 27$

Step 3

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

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Step 3.1

Replace all occurrences of $x$ in $y = x^{2} + x - 27$ with $5$ .

$y = {(5)}^{2} + 5 - 27$

$x = 5$

Step 3.2

Simplify $y = {(5)}^{2} + 5 - 27$ .

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Remove parentheses.

$y = {(5)}^{2} + 5 - 27$

$x = 5$

$y = {(5)}^{2} + 5 - 27$

$x = 5$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify ${(5)}^{2} + 5 - 27$ .

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Step 3.2.2.1.1

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

$y = 25 + 5 - 27$

$x = 5$

Step 3.2.2.1.2

Add $25$ and $5$ .

$y = 30 - 27$

$x = 5$

Step 3.2.2.1.3

Subtract $27$ from $30$ .

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

Step 4

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

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Step 4.1

Replace all occurrences of $x$ in $y = x^{2} + x - 27$ with $- 4$ .

$y = {(- 4)}^{2} - 4 - 27$

$x = - 4$

Step 4.2

Simplify $y = {(- 4)}^{2} - 4 - 27$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$y = {(- 4)}^{2} - 4 - 27$

$x = - 4$

$y = {(- 4)}^{2} - 4 - 27$

$x = - 4$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify ${(- 4)}^{2} - 4 - 27$ .

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Step 4.2.2.1.1

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

$y = 16 - 4 - 27$

$x = - 4$

Step 4.2.2.1.2

Subtract $4$ from $16$ .

$y = 12 - 27$

$x = - 4$

Step 4.2.2.1.3

Subtract $27$ from $12$ .

$y = - 15$

$x = - 4$

$y = - 15$

$x = - 4$

$y = - 15$

$x = - 4$

$y = - 15$

$x = - 4$

$y = - 15$

$x = - 4$

Step 5

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

$(5, 3)$

$(- 4, - 15)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(5, 3), (- 4, - 15)$

Equation Form:

$x = 5, y = 3$

$x = - 4, y = - 15$

Step 7