Pre-Algebra Examples

$\frac{1}{2 z} + 7 = 2 z + 13$

Step 1

Move all terms not containing $z$ to the right side of the equation.

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Step 1.1

Subtract $7$ from both sides of the equation.

$\frac{1}{2 z} = 2 z + 13 - 7$

Step 1.2

Subtract $7$ from $13$ .

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

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 z, 1, 1$

Step 2.2

The LCM of one and any expression is the expression.

$2 z$

Step 3

Multiply each term in $\frac{1}{2 z} = 2 z + 6$ by $2 z$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{1}{2 z} = 2 z + 6$ by $2 z$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Rewrite using the commutative property of multiplication.

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

Step 3.2.2

Cancel the common factor of $2$ .

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Step 3.2.2.1

Factor $2$ out of $2 z$ .

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

Step 3.2.2.2

Cancel the common factor.

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

Step 3.2.2.3

Rewrite the expression.

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

Step 3.2.3

Cancel the common factor of $z$ .

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Step 3.2.3.1

Cancel the common factor.

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

Step 3.2.3.2

Rewrite the expression.

$1 = 2 z (2 z) + 6 (2 z)$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Rewrite using the commutative property of multiplication.

$1 = 2 \cdot 2 z \cdot z + 6 (2 z)$

Step 3.3.1.2

Multiply $z$ by $z$ by adding the exponents.

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Step 3.3.1.2.1

Move $z$ .

$1 = 2 \cdot 2 (z \cdot z) + 6 (2 z)$

Step 3.3.1.2.2

Multiply $z$ by $z$ .

$1 = 2 \cdot 2 z^{2} + 6 (2 z)$

Step 3.3.1.3

Multiply $2$ by $2$ .

$1 = 4 z^{2} + 6 (2 z)$

Step 3.3.1.4

Multiply $2$ by $6$ .

$1 = 4 z^{2} + 12 z$

Step 4

Solve the equation.

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Step 4.1

Rewrite the equation as $4 z^{2} + 12 z = 1$ .

$4 z^{2} + 12 z = 1$

Step 4.2

Subtract $1$ from both sides of the equation.

$4 z^{2} + 12 z - 1 = 0$

Step 4.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4

Substitute the values $a = 4$ , $b = 12$ , and $c = - 1$ into the quadratic formula and solve for $z$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (4 \cdot - 1)}}{2 \cdot 4}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $12$ to the power of $2$ .

$z = \frac{- 12 \pm \sqrt{144 - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}$

Step 4.5.1.2

Multiply $- 4 \cdot 4 \cdot - 1$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $4$ .

$z = \frac{- 12 \pm \sqrt{144 - 16 \cdot - 1}}{2 \cdot 4}$

Step 4.5.1.2.2

Multiply $- 16$ by $- 1$ .

$z = \frac{- 12 \pm \sqrt{144 + 16}}{2 \cdot 4}$

Step 4.5.1.3

Add $144$ and $16$ .

$z = \frac{- 12 \pm \sqrt{160}}{2 \cdot 4}$

Step 4.5.1.4

Rewrite $160$ as $4^{2} \cdot 10$ .

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Step 4.5.1.4.1

Factor $16$ out of $160$ .

$z = \frac{- 12 \pm \sqrt{16 (10)}}{2 \cdot 4}$

Step 4.5.1.4.2

Rewrite $16$ as $4^{2}$ .

$z = \frac{- 12 \pm \sqrt{4^{2} \cdot 10}}{2 \cdot 4}$

Step 4.5.1.5

Pull terms out from under the radical.

$z = \frac{- 12 \pm 4 \sqrt{10}}{2 \cdot 4}$

Step 4.5.2

Multiply $2$ by $4$ .

$z = \frac{- 12 \pm 4 \sqrt{10}}{8}$

Step 4.5.3

Simplify $\frac{- 12 \pm 4 \sqrt{10}}{8}$ .

$z = \frac{- 3 \pm \sqrt{10}}{2}$

Step 4.6

The final answer is the combination of both solutions.

$z = - \frac{3 - \sqrt{10}}{2}, - \frac{3 + \sqrt{10}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$z = - \frac{3 - \sqrt{10}}{2}, - \frac{3 + \sqrt{10}}{2}$

Decimal Form:

$z = 0.08113883 \dots, - 3.08113883 \dots$