Algebra Examples

2 x + 3 = 3 x - 4

$2 x + 3 = 3 x - 4$

Step 1

Move all terms containing

x

$x$ to the left side of the equation.

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Step 1.1

Subtract

3 x

$3 x$ from both sides of the equation.

2 x + 3 - 3 x = - 4

$2 x + 3 - 3 x = - 4$

Step 1.2

Subtract

3 x

$3 x$ from

2 x

$2 x$ .

- x + 3 = - 4

$- x + 3 = - 4$

- x + 3 = - 4

$- x + 3 = - 4$

Step 2

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 2.1

Subtract

3

$3$ from both sides of the equation.

- x = - 4 - 3

$- x = - 4 - 3$

Step 2.2

Subtract

3

$3$ from

- 4

$- 4$ .

- x = - 7

$- x = - 7$

- x = - 7

$- x = - 7$

Step 3

Divide each term in

- x = - 7

$- x = - 7$ by

- 1

$- 1$ and simplify.

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Step 3.1

Divide each term in

- x = - 7

$- x = - 7$ by

- 1

$- 1$ .

\frac{- x}{- 1} = \frac{- 7}{- 1}

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

\frac{x}{1} = \frac{- 7}{- 1}

$\frac{x}{1} = \frac{- 7}{- 1}$

Step 3.2.2

Divide

x

$x$ by

1

$1$ .

x = \frac{- 7}{- 1}

$x = \frac{- 7}{- 1}$

x = \frac{- 7}{- 1}

$x = \frac{- 7}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide

$- 7$ by

$- 1$ .

$x = 7$

$x = 7$

$x = 7$

Enter a problem...

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