Pre-Algebra Examples

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

Step 1

Multiply both sides by $10$ .

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

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Cancel the common factor of $10$ .

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Step 2.1.1.1

Cancel the common factor.

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

Step 2.1.1.2

Rewrite the expression.

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

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $(\frac{5 x + 2}{4} - \frac{5}{4}) \cdot 10$ .

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Step 2.2.1.1

Combine the numerators over the common denominator.

$4 x + 1 = \frac{5 x + 2 - 5}{4} \cdot 10$

Step 2.2.1.2

Subtract $5$ from $2$ .

$4 x + 1 = \frac{5 x - 3}{4} \cdot 10$

Step 2.2.1.3

Cancel the common factor of $2$ .

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Step 2.2.1.3.1

Factor $2$ out of $4$ .

$4 x + 1 = \frac{5 x - 3}{2 (2)} \cdot 10$

Step 2.2.1.3.2

Factor $2$ out of $10$ .

$4 x + 1 = \frac{5 x - 3}{2 \cdot 2} \cdot (2 \cdot 5)$

Step 2.2.1.3.3

Cancel the common factor.

$4 x + 1 = \frac{5 x - 3}{2 \cdot 2} \cdot (2 \cdot 5)$

Step 2.2.1.3.4

Rewrite the expression.

$4 x + 1 = \frac{5 x - 3}{2} \cdot 5$

Step 2.2.1.4

Combine $\frac{5 x - 3}{2}$ and $5$ .

$4 x + 1 = \frac{(5 x - 3) \cdot 5}{2}$

Step 2.2.1.5

Move $5$ to the left of $5 x - 3$ .

$4 x + 1 = \frac{5 (5 x - 3)}{2}$

Step 3

Solve for $x$ .

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Step 3.1

Multiply both sides by $2$ .

$(4 x + 1) \cdot 2 = \frac{5 (5 x - 3)}{2} \cdot 2$

Step 3.2

Simplify.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $(4 x + 1) \cdot 2$ .

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Step 3.2.1.1.1

Apply the distributive property.

$4 x \cdot 2 + 1 \cdot 2 = \frac{5 (5 x - 3)}{2} \cdot 2$

Step 3.2.1.1.2

Multiply.

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Step 3.2.1.1.2.1

Multiply $2$ by $4$ .

$8 x + 1 \cdot 2 = \frac{5 (5 x - 3)}{2} \cdot 2$

Step 3.2.1.1.2.2

Multiply $2$ by $1$ .

$8 x + 2 = \frac{5 (5 x - 3)}{2} \cdot 2$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $\frac{5 (5 x - 3)}{2} \cdot 2$ .

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Step 3.2.2.1.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1.1

Cancel the common factor.

$8 x + 2 = \frac{5 (5 x - 3)}{2} \cdot 2$

Step 3.2.2.1.1.2

Rewrite the expression.

$8 x + 2 = 5 (5 x - 3)$

Step 3.2.2.1.2

Apply the distributive property.

$8 x + 2 = 5 (5 x) + 5 \cdot - 3$

Step 3.2.2.1.3

Multiply.

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Step 3.2.2.1.3.1

Multiply $5$ by $5$ .

$8 x + 2 = 25 x + 5 \cdot - 3$

Step 3.2.2.1.3.2

Multiply $5$ by $- 3$ .

$8 x + 2 = 25 x - 15$

Step 3.3

Solve for $x$ .

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Step 3.3.1

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

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Step 3.3.1.1

Subtract $25 x$ from both sides of the equation.

$8 x + 2 - 25 x = - 15$

Step 3.3.1.2

Subtract $25 x$ from $8 x$ .

$- 17 x + 2 = - 15$

Step 3.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 3.3.2.1

Subtract $2$ from both sides of the equation.

$- 17 x = - 15 - 2$

Step 3.3.2.2

Subtract $2$ from $- 15$ .

$- 17 x = - 17$

Step 3.3.3

Divide each term in $- 17 x = - 17$ by $- 17$ and simplify.

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Step 3.3.3.1

Divide each term in $- 17 x = - 17$ by $- 17$ .

$\frac{- 17 x}{- 17} = \frac{- 17}{- 17}$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 17$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 17 x}{- 17} = \frac{- 17}{- 17}$

Step 3.3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 17}{- 17}$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $- 17$ by $- 17$ .

$x = 1$