Algebra Examples

\frac{2 + 6 x}{2} = 3 x + 3

$\frac{2 + 6 x}{2} = 3 x + 3$

Step 1

Cancel the common factor of

2 + 6 x

$2 + 6 x$ and

2

$2$ .

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Step 1.1

Factor

2

$2$ out of

2

$2$ .

\frac{2 \cdot 1 + 6 x}{2} = 3 x + 3

$\frac{2 \cdot 1 + 6 x}{2} = 3 x + 3$

Step 1.2

Factor

2

$2$ out of

6 x

$6 x$ .

\frac{2 \cdot 1 + 2 (3 x)}{2} = 3 x + 3

$\frac{2 \cdot 1 + 2 (3 x)}{2} = 3 x + 3$

Step 1.3

Factor

2

$2$ out of

2 (1) + 2 (3 x)

$2 (1) + 2 (3 x)$ .

\frac{2 (1 + 3 x)}{2} = 3 x + 3

$\frac{2 (1 + 3 x)}{2} = 3 x + 3$

Step 1.4

Cancel the common factors.

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Step 1.4.1

Factor

2

$2$ out of

2

$2$ .

\frac{2 (1 + 3 x)}{2 (1)} = 3 x + 3

$\frac{2 (1 + 3 x)}{2 (1)} = 3 x + 3$

Step 1.4.2

Cancel the common factor.

$\frac{2 (1 + 3 x)}{2 \cdot 1} = 3 x + 3$

Step 1.4.3

Rewrite the expression.

$\frac{1 + 3 x}{1} = 3 x + 3$

Step 1.4.4

Divide

$1 + 3 x$ by

$1$ .

$1 + 3 x = 3 x + 3$

$1 + 3 x = 3 x + 3$

$1 + 3 x = 3 x + 3$

Step 2

Solve for

$x$ .

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Step 2.1

Move all terms containing

$x$ to the left side of the equation.

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Step 2.1.1

Subtract

$3 x$ from both sides of the equation.

$1 + 3 x - 3 x = 3$

Step 2.1.2

Combine the opposite terms in

$1 + 3 x - 3 x$ .

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Step 2.1.2.1

Subtract

$3 x$ from

$3 x$ .

$1 + 0 = 3$

Step 2.1.2.2

Add

$1$ and

$0$ .

$1 = 3$

$1 = 3$

$1 = 3$

Step 2.2

Since

$1 \neq 3$ , there are no solutions.

No solution

No solution

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