Algebra Examples

$\frac{1}{5 x} - \frac{1}{4 x} + \frac{1}{3 x} = - \frac{17}{60}$

Step 1

Add $\frac{17}{60}$ to both sides of the equation.

$\frac{1}{5 x} - \frac{1}{4 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2

Simplify $\frac{1}{5 x} - \frac{1}{4 x} + \frac{1}{3 x} + \frac{17}{60}$ .

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Step 2.1

To write $\frac{1}{5 x}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{1}{5 x} \cdot \frac{4}{4} - \frac{1}{4 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.2

To write $- \frac{1}{4 x}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{1}{5 x} \cdot \frac{4}{4} - \frac{1}{4 x} \cdot \frac{5}{5} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.3

Write each expression with a common denominator of $20 x$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{1}{5 x}$ by $\frac{4}{4}$ .

$\frac{4}{5 x \cdot 4} - \frac{1}{4 x} \cdot \frac{5}{5} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.3.2

Multiply $4$ by $5$ .

$\frac{4}{20 x} - \frac{1}{4 x} \cdot \frac{5}{5} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.3.3

Multiply $\frac{1}{4 x}$ by $\frac{5}{5}$ .

$\frac{4}{20 x} - \frac{5}{4 x \cdot 5} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.3.4

Multiply $5$ by $4$ .

$\frac{4}{20 x} - \frac{5}{20 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{4 - 5}{20 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.5

Subtract $5$ from $4$ .

$\frac{- 1}{20 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.6

Move the negative in front of the fraction.

$- \frac{1}{20 x} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.7

To write $- \frac{1}{20 x}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{1}{20 x} \cdot \frac{3}{3} + \frac{1}{3 x} + \frac{17}{60} = 0$

Step 2.8

To write $\frac{1}{3 x}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$- \frac{1}{20 x} \cdot \frac{3}{3} + \frac{1}{3 x} \cdot \frac{20}{20} + \frac{17}{60} = 0$

Step 2.9

Write each expression with a common denominator of $60 x$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.9.1

Multiply $\frac{1}{20 x}$ by $\frac{3}{3}$ .

$- \frac{3}{20 x \cdot 3} + \frac{1}{3 x} \cdot \frac{20}{20} + \frac{17}{60} = 0$

Step 2.9.2

Multiply $3$ by $20$ .

$- \frac{3}{60 x} + \frac{1}{3 x} \cdot \frac{20}{20} + \frac{17}{60} = 0$

Step 2.9.3

Multiply $\frac{1}{3 x}$ by $\frac{20}{20}$ .

$- \frac{3}{60 x} + \frac{20}{3 x \cdot 20} + \frac{17}{60} = 0$

Step 2.9.4

Multiply $20$ by $3$ .

$- \frac{3}{60 x} + \frac{20}{60 x} + \frac{17}{60} = 0$

Step 2.10

Combine the numerators over the common denominator.

$\frac{- 3 + 20}{60 x} + \frac{17}{60} = 0$

Step 2.11

Add $- 3$ and $20$ .

$\frac{17}{60 x} + \frac{17}{60} = 0$

Step 3

To write $\frac{17}{60}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{17}{60 x} + \frac{17}{60} \cdot \frac{x}{x} = 0$

Step 4

Multiply $\frac{17}{60}$ by $\frac{x}{x}$ .

$\frac{17}{60 x} + \frac{17 x}{60 x} = 0$

Step 5

Combine the numerators over the common denominator.

$\frac{17 + 17 x}{60 x} = 0$

Step 6

Factor $17$ out of $17 + 17 x$ .

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Step 6.1

Factor $17$ out of $17$ .

$\frac{17 \cdot 1 + 17 x}{60 x} = 0$

Step 6.2

Factor $17$ out of $17 \cdot 1 + 17 x$ .

$\frac{17 \cdot (1 + x)}{60 x} = 0$

Step 6.3

Multiply $17$ by $1 + x$ .

$\frac{17 (1 + x)}{60 x} = 0$

Step 7

Set the numerator equal to zero.

$17 (1 + x) = 0$

Step 8

Solve the equation for $x$ .

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Step 8.1

Divide each term in $17 (1 + x) = 0$ by $17$ and simplify.

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Step 8.1.1

Divide each term in $17 (1 + x) = 0$ by $17$ .

$\frac{17 (1 + x)}{17} = \frac{0}{17}$

Step 8.1.2

Simplify the left side.

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Step 8.1.2.1

Cancel the common factor of $17$ .

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Step 8.1.2.1.1

Cancel the common factor.

$\frac{17 (1 + x)}{17} = \frac{0}{17}$

Step 8.1.2.1.2

Divide $1 + x$ by $1$ .

$1 + x = \frac{0}{17}$

Step 8.1.3

Simplify the right side.

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Step 8.1.3.1

Divide $0$ by $17$ .

$1 + x = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$x = - 1$