Algebra Examples

$\frac{10 x - 4}{2} + 7 = 5 (x + 1)$

Step 1

Simplify $\frac{10 x - 4}{2} + 7$ .

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Step 1.1

Cancel the common factor of $10 x - 4$ and $2$ .

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Step 1.1.1

Factor $2$ out of $10 x$ .

$\frac{2 (5 x) - 4}{2} + 7 = 5 (x + 1)$

Step 1.1.2

Factor $2$ out of $- 4$ .

$\frac{2 (5 x) + 2 \cdot - 2}{2} + 7 = 5 (x + 1)$

Step 1.1.3

Factor $2$ out of $2 (5 x) + 2 (- 2)$ .

$\frac{2 (5 x - 2)}{2} + 7 = 5 (x + 1)$

Step 1.1.4

Cancel the common factors.

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Step 1.1.4.1

Factor $2$ out of $2$ .

$\frac{2 (5 x - 2)}{2 (1)} + 7 = 5 (x + 1)$

Step 1.1.4.2

Cancel the common factor.

$\frac{2 (5 x - 2)}{2 \cdot 1} + 7 = 5 (x + 1)$

Step 1.1.4.3

Rewrite the expression.

$\frac{5 x - 2}{1} + 7 = 5 (x + 1)$

Step 1.1.4.4

Divide $5 x - 2$ by $1$ .

$5 x - 2 + 7 = 5 (x + 1)$

$5 x - 2 + 7 = 5 (x + 1)$

$5 x - 2 + 7 = 5 (x + 1)$

Step 1.2

Add $- 2$ and $7$ .

$5 x + 5 = 5 (x + 1)$

$5 x + 5 = 5 (x + 1)$

Step 2

Simplify $5 (x + 1)$ .

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Step 2.1

Apply the distributive property.

$5 x + 5 = 5 x + 5 \cdot 1$

Step 2.2

Multiply $5$ by $1$ .

$5 x + 5 = 5 x + 5$

$5 x + 5 = 5 x + 5$

Step 3

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

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Step 3.1

Subtract $5 x$ from both sides of the equation.

$5 x + 5 - 5 x = 5$

Step 3.2

Combine the opposite terms in $5 x + 5 - 5 x$ .

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Step 3.2.1

Subtract $5 x$ from $5 x$ .

$0 + 5 = 5$

Step 3.2.2

Add $0$ and $5$ .

$5 = 5$

$5 = 5$

$5 = 5$

Step 4

Since $5 = 5$ , the equation will always be true for any value of $x$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$