Algebra Examples

$y = 9 - x$ $y = 2 x^{2} + 4 x + 6$

Step 1

Eliminate the equal sides of each equation and combine.

$9 - x = 2 x^{2} + 4 x + 6$

Step 2

Solve $9 - x = 2 x^{2} + 4 x + 6$ for $x$ .

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Step 2.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x^{2} + 4 x + 6 = 9 - x$

Step 2.2

Move all terms containing $x$ to the left side of the equation.

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Step 2.2.1

Add $x$ to both sides of the equation.

$2 x^{2} + 4 x + 6 + x = 9$

Step 2.2.2

Add $4 x$ and $x$ .

$2 x^{2} + 5 x + 6 = 9$

Step 2.3

Subtract $9$ from both sides of the equation.

$2 x^{2} + 5 x + 6 - 9 = 0$

Step 2.4

Subtract $9$ from $6$ .

$2 x^{2} + 5 x - 3 = 0$

Step 2.5

Factor by grouping.

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Step 2.5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

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Step 2.5.1.1

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 3 = 0$

Step 2.5.1.2

Rewrite $5$ as $- 1$ plus $6$

$2 x^{2} + (- 1 + 6) x - 3 = 0$

Step 2.5.1.3

Apply the distributive property.

$2 x^{2} - 1 x + 6 x - 3 = 0$

Step 2.5.2

Factor out the greatest common factor from each group.

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Step 2.5.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 6 x - 3 = 0$

Step 2.5.2.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x - 1) + 3 (2 x - 1) = 0$

Step 2.5.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 3) = 0$

Step 2.6

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

$2 x - 1 = 0$

$x + 3 = 0 + y = 2 x^{2} + 4 x + 6$

Step 2.7

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.7.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.7.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.7.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.7.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.7.2.2.2

Simplify the left side.

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Step 2.7.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.7.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.7.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 2.8

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

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Step 2.8.1

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

$x + 3 = 0$

Step 2.8.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.9

The final solution is all the values that make $(2 x - 1) (x + 3) = 0$ true.

$x = \frac{1}{2}, - 3$

Step 3

Evaluate $y$ when $x = \frac{1}{2}$ .

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Step 3.1

Substitute $\frac{1}{2}$ for $x$ .

$y = 2 {(\frac{1}{2})}^{2} + 4 (\frac{1}{2}) + 6$

Step 3.2

Simplify $2 {(\frac{1}{2})}^{2} + 4 (\frac{1}{2}) + 6$ .

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to $\frac{1}{2}$ .

$y = 2 \frac{1^{2}}{2^{2}} + 4 (\frac{1}{2}) + 6$

Step 3.2.1.2

One to any power is one.

$y = 2 \frac{1}{2^{2}} + 4 (\frac{1}{2}) + 6$

Step 3.2.1.3

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

$y = 2 (\frac{1}{4}) + 4 (\frac{1}{2}) + 6$

Step 3.2.1.4

Cancel the common factor of $2$ .

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Step 3.2.1.4.1

Factor $2$ out of $4$ .

$y = 2 \frac{1}{2 (2)} + 4 (\frac{1}{2}) + 6$

Step 3.2.1.4.2

Cancel the common factor.

$y = 2 \frac{1}{2 \cdot 2} + 4 (\frac{1}{2}) + 6$

Step 3.2.1.4.3

Rewrite the expression.

$y = \frac{1}{2} + 4 (\frac{1}{2}) + 6$

Step 3.2.1.5

Cancel the common factor of $2$ .

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Step 3.2.1.5.1

Factor $2$ out of $4$ .

$y = \frac{1}{2} + 2 (2) \frac{1}{2} + 6$

Step 3.2.1.5.2

Cancel the common factor.

$y = \frac{1}{2} + 2 \cdot 2 \frac{1}{2} + 6$

Step 3.2.1.5.3

Rewrite the expression.

$y = \frac{1}{2} + 2 + 6$

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Write $2$ as a fraction with denominator $1$ .

$y = \frac{1}{2} + \frac{2}{1} + 6$

Step 3.2.2.2

Multiply $\frac{2}{1}$ by $\frac{2}{2}$ .

$y = \frac{1}{2} + \frac{2}{1} \cdot \frac{2}{2} + 6$

Step 3.2.2.3

Multiply $\frac{2}{1}$ by $\frac{2}{2}$ .

$y = \frac{1}{2} + \frac{2 \cdot 2}{2} + 6$

Step 3.2.2.4

Write $6$ as a fraction with denominator $1$ .

$y = \frac{1}{2} + \frac{2 \cdot 2}{2} + \frac{6}{1}$

Step 3.2.2.5

Multiply $\frac{6}{1}$ by $\frac{2}{2}$ .

$y = \frac{1}{2} + \frac{2 \cdot 2}{2} + \frac{6}{1} \cdot \frac{2}{2}$

Step 3.2.2.6

Multiply $\frac{6}{1}$ by $\frac{2}{2}$ .

$y = \frac{1}{2} + \frac{2 \cdot 2}{2} + \frac{6 \cdot 2}{2}$

Step 3.2.3

Combine the numerators over the common denominator.

$y = \frac{1 + 2 \cdot 2 + 6 \cdot 2}{2}$

Step 3.2.4

Simplify each term.

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Step 3.2.4.1

Multiply $2$ by $2$ .

$y = \frac{1 + 4 + 6 \cdot 2}{2}$

Step 3.2.4.2

Multiply $6$ by $2$ .

$y = \frac{1 + 4 + 12}{2}$

Step 3.2.5

Simplify by adding numbers.

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Step 3.2.5.1

Add $1$ and $4$ .

$y = \frac{5 + 12}{2}$

Step 3.2.5.2

Add $5$ and $12$ .

$y = \frac{17}{2}$

Step 4

Evaluate $y$ when $x = - 3$ .

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Step 4.1

Substitute $- 3$ for $x$ .

$y = 2 {(- 3)}^{2} + 4 (- 3) + 6$

Step 4.2

Simplify $2 {(- 3)}^{2} + 4 (- 3) + 6$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

$y = 2 \cdot 9 + 4 (- 3) + 6$

Step 4.2.1.2

Multiply $2$ by $9$ .

$y = 18 + 4 (- 3) + 6$

Step 4.2.1.3

Multiply $4$ by $- 3$ .

$y = 18 - 12 + 6$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Subtract $12$ from $18$ .

$y = 6 + 6$

Step 4.2.2.2

Add $6$ and $6$ .

$y = 12$

Step 5

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

$(\frac{1}{2}, \frac{17}{2})$

$(- 3, 12)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{2}, \frac{17}{2}), (- 3, 12)$

Equation Form:

$x = \frac{1}{2}, y = \frac{17}{2}$

$x = - 3, y = 12$

Step 7