Basic Math Examples

$\frac{2}{5} h - 7 = \frac{12}{5} h - 2 h + 3$

Step 1

Since $h$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\frac{12}{5} h - 2 h + 3 = \frac{2}{5} h - 7$

Step 2

Simplify $\frac{12}{5} h - 2 h + 3$ .

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Step 2.1

Combine $\frac{12}{5}$ and $h$ .

$\frac{12 h}{5} - 2 h + 3 = \frac{2}{5} h - 7$

Step 2.2

To write $- 2 h$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{12 h}{5} - 2 h \cdot \frac{5}{5} + 3 = \frac{2}{5} h - 7$

Step 2.3

Simplify terms.

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Step 2.3.1

Combine $- 2 h$ and $\frac{5}{5}$ .

$\frac{12 h}{5} + \frac{- 2 h \cdot 5}{5} + 3 = \frac{2}{5} h - 7$

Step 2.3.2

Combine the numerators over the common denominator.

$\frac{12 h - 2 h \cdot 5}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Factor $2 h$ out of $12 h - 2 h \cdot 5$ .

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Step 2.4.1.1

Factor $2 h$ out of $12 h$ .

$\frac{2 h (6) - 2 h \cdot 5}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4.1.2

Factor $2 h$ out of $- 2 h \cdot 5$ .

$\frac{2 h (6) + 2 h (- 1 \cdot 5)}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4.1.3

Factor $2 h$ out of $2 h (6) + 2 h (- 1 \cdot 5)$ .

$\frac{2 h (6 - 1 \cdot 5)}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4.2

Multiply $- 1$ by $5$ .

$\frac{2 h (6 - 5)}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4.3

Subtract $5$ from $6$ .

$\frac{2 h \cdot 1}{5} + 3 = \frac{2}{5} h - 7$

Step 2.4.4

Multiply $2$ by $1$ .

$\frac{2 h}{5} + 3 = \frac{2}{5} h - 7$

Step 3

Combine $\frac{2}{5}$ and $h$ .

$\frac{2 h}{5} + 3 = \frac{2 h}{5} - 7$

Step 4

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

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Step 4.1

Subtract $\frac{2 h}{5}$ from both sides of the equation.

$\frac{2 h}{5} + 3 - \frac{2 h}{5} = - 7$

Step 4.2

Combine the opposite terms in $\frac{2 h}{5} + 3 - \frac{2 h}{5}$ .

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Step 4.2.1

Subtract $\frac{2 h}{5}$ from $\frac{2 h}{5}$ .

$0 + 3 = - 7$

Step 4.2.2

Add $0$ and $3$ .

$3 = - 7$

Step 5

Since $3 \neq - 7$ , there are no solutions.

No solution