Basic Math Examples

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{4 z + 4}{10}$

Step 1

Cancel the common factor of $4 z + 4$ and $10$ .

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Step 1.1

Factor $2$ out of $4 z$ .

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 (2 z) + 4}{10}$

Step 1.2

Factor $2$ out of $4$ .

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 (2 z) + 2 (2)}{10}$

Step 1.3

Factor $2$ out of $2 (2 z) + 2 (2)$ .

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 (2 z + 2)}{10}$

Step 1.4

Cancel the common factors.

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Step 1.4.1

Factor $2$ out of $10$ .

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 (2 z + 2)}{2 (5)}$

Step 1.4.2

Cancel the common factor.

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 (2 z + 2)}{2 \cdot 5}$

Step 1.4.3

Rewrite the expression.

$\frac{2 z - 6}{5} + \frac{16}{10} = \frac{2 z + 2}{5}$

Step 2

Multiply both sides by $5$ .

$(\frac{2 z - 6}{5} + \frac{16}{10}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3

Simplify.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $(\frac{2 z - 6}{5} + \frac{16}{10}) \cdot 5$ .

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Step 3.1.1.1

Simplify each term.

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Step 3.1.1.1.1

Factor $2$ out of $2 z - 6$ .

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Step 3.1.1.1.1.1

Factor $2$ out of $2 z$ .

$(\frac{2 (z) - 6}{5} + \frac{16}{10}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.1.2

Factor $2$ out of $- 6$ .

$(\frac{2 z + 2 \cdot - 3}{5} + \frac{16}{10}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.1.3

Factor $2$ out of $2 z + 2 \cdot - 3$ .

$(\frac{2 (z - 3)}{5} + \frac{16}{10}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.2

Cancel the common factor of $16$ and $10$ .

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Step 3.1.1.1.2.1

Factor $2$ out of $16$ .

$(\frac{2 (z - 3)}{5} + \frac{2 (8)}{10}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.2.2

Cancel the common factors.

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Step 3.1.1.1.2.2.1

Factor $2$ out of $10$ .

$(\frac{2 (z - 3)}{5} + \frac{2 \cdot 8}{2 \cdot 5}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.2.2.2

Cancel the common factor.

$(\frac{2 (z - 3)}{5} + \frac{2 \cdot 8}{2 \cdot 5}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.1.2.2.3

Rewrite the expression.

$(\frac{2 (z - 3)}{5} + \frac{8}{5}) \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.2

Combine the numerators over the common denominator.

$\frac{2 (z - 3) + 8}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.3

Simplify each term.

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Step 3.1.1.3.1

Apply the distributive property.

$\frac{2 z + 2 \cdot - 3 + 8}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.3.2

Multiply $2$ by $- 3$ .

$\frac{2 z - 6 + 8}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4

Simplify terms.

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Step 3.1.1.4.1

Add $- 6$ and $8$ .

$\frac{2 z + 2}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.2

Factor $2$ out of $2 z + 2$ .

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Step 3.1.1.4.2.1

Factor $2$ out of $2 z$ .

$\frac{2 (z) + 2}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.2.2

Factor $2$ out of $2$ .

$\frac{2 (z) + 2 (1)}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.2.3

Factor $2$ out of $2 (z) + 2 (1)$ .

$\frac{2 (z + 1)}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.3

Cancel the common factor of $5$ .

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Step 3.1.1.4.3.1

Cancel the common factor.

$\frac{2 (z + 1)}{5} \cdot 5 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.3.2

Rewrite the expression.

$2 (z + 1) = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.4

Apply the distributive property.

$2 z + 2 \cdot 1 = \frac{2 z + 2}{5} \cdot 5$

Step 3.1.1.4.5

Multiply $2$ by $1$ .

$2 z + 2 = \frac{2 z + 2}{5} \cdot 5$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\frac{2 z + 2}{5} \cdot 5$ .

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Step 3.2.1.1

Factor $2$ out of $2 z + 2$ .

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Step 3.2.1.1.1

Factor $2$ out of $2 z$ .

$2 z + 2 = \frac{2 (z) + 2}{5} \cdot 5$

Step 3.2.1.1.2

Factor $2$ out of $2$ .

$2 z + 2 = \frac{2 (z) + 2 (1)}{5} \cdot 5$

Step 3.2.1.1.3

Factor $2$ out of $2 (z) + 2 (1)$ .

$2 z + 2 = \frac{2 (z + 1)}{5} \cdot 5$

Step 3.2.1.2

Cancel the common factor of $5$ .

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Step 3.2.1.2.1

Cancel the common factor.

$2 z + 2 = \frac{2 (z + 1)}{5} \cdot 5$

Step 3.2.1.2.2

Rewrite the expression.

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

Step 3.2.1.3

Apply the distributive property.

$2 z + 2 = 2 z + 2 \cdot 1$

Step 3.2.1.4

Multiply $2$ by $1$ .

$2 z + 2 = 2 z + 2$

Step 4

Solve for $z$ .

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Step 4.1

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

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Step 4.1.1

Subtract $2 z$ from both sides of the equation.

$2 z + 2 - 2 z = 2$

Step 4.1.2

Combine the opposite terms in $2 z + 2 - 2 z$ .

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Step 4.1.2.1

Subtract $2 z$ from $2 z$ .

$0 + 2 = 2$

Step 4.1.2.2

Add $0$ and $2$ .

$2 = 2$

Step 4.2

Since $2 = 2$ , the equation will always be true for any value of $z$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$