Algebra Examples

$\frac{1}{2} (8 x + 26) = 13 + 4 x$

Step 1

Simplify $\frac{1}{2} \cdot (8 x + 26)$ .

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Step 1.1

Rewrite.

$0 + 0 + \frac{1}{2} \cdot (8 x + 26) = 13 + 4 x$

Step 1.2

Simplify by adding zeros.

$\frac{1}{2} \cdot (8 x + 26) = 13 + 4 x$

Step 1.3

Apply the distributive property.

$\frac{1}{2} (8 x) + \frac{1}{2} \cdot 26 = 13 + 4 x$

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Factor $2$ out of $8 x$ .

$\frac{1}{2} (2 (4 x)) + \frac{1}{2} \cdot 26 = 13 + 4 x$

Step 1.4.2

Cancel the common factor.

$\frac{1}{2} (2 (4 x)) + \frac{1}{2} \cdot 26 = 13 + 4 x$

Step 1.4.3

Rewrite the expression.

$4 x + \frac{1}{2} \cdot 26 = 13 + 4 x$

Step 1.5

Cancel the common factor of $2$ .

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Step 1.5.1

Factor $2$ out of $26$ .

$4 x + \frac{1}{2} \cdot (2 (13)) = 13 + 4 x$

Step 1.5.2

Cancel the common factor.

$4 x + \frac{1}{2} \cdot (2 \cdot 13) = 13 + 4 x$

Step 1.5.3

Rewrite the expression.

$4 x + 13 = 13 + 4 x$

Step 2

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

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Step 2.1

Subtract $4 x$ from both sides of the equation.

$4 x + 13 - 4 x = 13$

Step 2.2

Combine the opposite terms in $4 x + 13 - 4 x$ .

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Step 2.2.1

Subtract $4 x$ from $4 x$ .

$0 + 13 = 13$

Step 2.2.2

Add $0$ and $13$ .

$13 = 13$

Step 3

Since $13 = 13$ , the equation will always be true for any value of $x$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$