Algebra Examples

$y + 12 = x^{2} + x$ $x + y = 3$

Step 1

Subtract $12$ from both sides of the equation.

$y = x^{2} + x - 12$

$x + y = 3$

Step 2

Replace all occurrences of $y$ with $x^{2} + x - 12$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} + x - 12$ .

$x + x^{2} + x - 12 = 3$

$y = x^{2} + x - 12$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $x + x^{2} + x - 12$ .

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Step 2.2.1.1

Remove parentheses.

$x + x^{2} + x - 12 = 3$

$y = x^{2} + x - 12$

Step 2.2.1.2

Add $x$ and $x$ .

$x^{2} + 2 x - 12 = 3$

$y = x^{2} + x - 12$

$x^{2} + 2 x - 12 = 3$

$y = x^{2} + x - 12$

$x^{2} + 2 x - 12 = 3$

$y = x^{2} + x - 12$

$x^{2} + 2 x - 12 = 3$

$y = x^{2} + x - 12$

Step 3

Solve for $x$ in $x^{2} + 2 x - 12 = 3$ .

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Step 3.1

Subtract $3$ from both sides of the equation.

$x^{2} + 2 x - 12 - 3 = 0$

$y = x^{2} + x - 12$

Step 3.2

Subtract $3$ from $- 12$ .

$x^{2} + 2 x - 15 = 0$

$y = x^{2} + x - 12$

Step 3.3

Factor $x^{2} + 2 x - 15$ using the AC method.

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Step 3.3.1

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

$y = x^{2} + x - 12$

Step 3.3.2

Write the factored form using these integers.

$(x - 3) (x + 5) = 0$

$y = x^{2} + x - 12$

$(x - 3) (x + 5) = 0$

$y = x^{2} + x - 12$

Step 3.4

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

$x - 3 = 0$

$x + 5 = 0$

$y = x^{2} + x - 12$

Step 3.5

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

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Step 3.5.1

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

$x - 3 = 0$

$y = x^{2} + x - 12$

Step 3.5.2

Add $3$ to both sides of the equation.

$x = 3$

$y = x^{2} + x - 12$

$x = 3$

$y = x^{2} + x - 12$

Step 3.6

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

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Step 3.6.1

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

$x + 5 = 0$

$y = x^{2} + x - 12$

Step 3.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = x^{2} + x - 12$

$x = - 5$

$y = x^{2} + x - 12$

Step 3.7

The final solution is all the values that make $(x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

$y = x^{2} + x - 12$

$x = 3, - 5$

$y = x^{2} + x - 12$

Step 4

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

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Step 4.1

Replace all occurrences of $x$ in $y = x^{2} + x - 12$ with $3$ .

$y = {(3)}^{2} + 3 - 12$

$x = 3$

Step 4.2

Simplify $y = {(3)}^{2} + 3 - 12$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$y = {(3)}^{2} + 3 - 12$

$x = 3$

$y = {(3)}^{2} + 3 - 12$

$x = 3$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify ${(3)}^{2} + 3 - 12$ .

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Step 4.2.2.1.1

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

$y = 9 + 3 - 12$

$x = 3$

Step 4.2.2.1.2

Add $9$ and $3$ .

$y = 12 - 12$

$x = 3$

Step 4.2.2.1.3

Subtract $12$ from $12$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 5

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

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Step 5.1

Replace all occurrences of $x$ in $y = x^{2} + x - 12$ with $- 5$ .

$y = {(- 5)}^{2} - 5 - 12$

$x = - 5$

Step 5.2

Simplify $y = {(- 5)}^{2} - 5 - 12$ .

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Remove parentheses.

$y = {(- 5)}^{2} - 5 - 12$

$x = - 5$

$y = {(- 5)}^{2} - 5 - 12$

$x = - 5$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify ${(- 5)}^{2} - 5 - 12$ .

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Step 5.2.2.1.1

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

$y = 25 - 5 - 12$

$x = - 5$

Step 5.2.2.1.2

Subtract $5$ from $25$ .

$y = 20 - 12$

$x = - 5$

Step 5.2.2.1.3

Subtract $12$ from $20$ .

$y = 8$

$x = - 5$

$y = 8$

$x = - 5$

$y = 8$

$x = - 5$

$y = 8$

$x = - 5$

$y = 8$

$x = - 5$

Step 6

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

$(3, 0)$

$(- 5, 8)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, 0), (- 5, 8)$

Equation Form:

$x = 3, y = 0$

$x = - 5, y = 8$

Step 8