Algebra Examples

$\frac{x}{3 x - 5} \leq \frac{2}{x - 1}$

Step 1

Subtract $\frac{2}{x - 1}$ from both sides of the inequality.

$\frac{x}{3 x - 5} - \frac{2}{x - 1} \leq 0$

Step 2

Simplify $\frac{x}{3 x - 5} - \frac{2}{x - 1}$ .

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Step 2.1

To write $\frac{x}{3 x - 5}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{x}{3 x - 5} \cdot \frac{x - 1}{x - 1} - \frac{2}{x - 1} \leq 0$

Step 2.2

To write $- \frac{2}{x - 1}$ as a fraction with a common denominator, multiply by $\frac{3 x - 5}{3 x - 5}$ .

$\frac{x}{3 x - 5} \cdot \frac{x - 1}{x - 1} - \frac{2}{x - 1} \cdot \frac{3 x - 5}{3 x - 5} \leq 0$

Step 2.3

Write each expression with a common denominator of $(3 x - 5) (x - 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{x}{3 x - 5}$ by $\frac{x - 1}{x - 1}$ .

$\frac{x (x - 1)}{(3 x - 5) (x - 1)} - \frac{2}{x - 1} \cdot \frac{3 x - 5}{3 x - 5} \leq 0$

Step 2.3.2

Multiply $\frac{2}{x - 1}$ by $\frac{3 x - 5}{3 x - 5}$ .

$\frac{x (x - 1)}{(3 x - 5) (x - 1)} - \frac{2 (3 x - 5)}{(x - 1) (3 x - 5)} \leq 0$

Step 2.3.3

Reorder the factors of $(x - 1) (3 x - 5)$ .

$\frac{x (x - 1)}{(3 x - 5) (x - 1)} - \frac{2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{x (x - 1) - 2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 1 - 2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.2

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 1 - 2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.3

Move $- 1$ to the left of $x$ .

$\frac{x^{2} - 1 \cdot x - 2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.4

Rewrite $- 1 x$ as $- x$ .

$\frac{x^{2} - x - 2 (3 x - 5)}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.5

Apply the distributive property.

$\frac{x^{2} - x - 2 (3 x) - 2 \cdot - 5}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.6

Multiply $3$ by $- 2$ .

$\frac{x^{2} - x - 6 x - 2 \cdot - 5}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.7

Multiply $- 2$ by $- 5$ .

$\frac{x^{2} - x - 6 x + 10}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.8

Subtract $6 x$ from $- x$ .

$\frac{x^{2} - 7 x + 10}{(3 x - 5) (x - 1)} \leq 0$

Step 2.5.9

Factor $x^{2} - 7 x + 10$ using the AC method.

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Step 2.5.9.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $- 7$ .

$- 5, - 2$

Step 2.5.9.2

Write the factored form using these integers.

$\frac{(x - 5) (x - 2)}{(3 x - 5) (x - 1)} \leq 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 5 = 0$

$x - 2 = 0$

$3 x - 5 = 0$

$x - 1 = 0$

Step 4

Add $5$ to both sides of the equation.

$x = 5$

Step 5

Add $2$ to both sides of the equation.

$x = 2$

Step 6

Add $5$ to both sides of the equation.

$3 x = 5$

Step 7

Divide each term in $3 x = 5$ by $3$ and simplify.

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Step 7.1

Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $3$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{3}$

Step 8

Add $1$ to both sides of the equation.

$x = 1$

Step 9

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 5$

$x = 2$

$x = \frac{5}{3}$

$x = 1$

Step 10

Consolidate the solutions.

$x = 5, 2, \frac{5}{3}, 1$

Step 11

Find the domain of $\frac{(x - 5) (x - 2)}{(3 x - 5) (x - 1)}$ .

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Step 11.1

Set the denominator in $\frac{(x - 5) (x - 2)}{(3 x - 5) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(3 x - 5) (x - 1) = 0$

Step 11.2

Solve for $x$ .

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Step 11.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 5 = 0$

$x - 1 = 0$

Step 11.2.2

Set $3 x - 5$ equal to $0$ and solve for $x$ .

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Step 11.2.2.1

Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Step 11.2.2.2

Solve $3 x - 5 = 0$ for $x$ .

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Step 11.2.2.2.1

Add $5$ to both sides of the equation.

$3 x = 5$

Step 11.2.2.2.2

Divide each term in $3 x = 5$ by $3$ and simplify.

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Step 11.2.2.2.2.1

Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Step 11.2.2.2.2.2

Simplify the left side.

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Step 11.2.2.2.2.2.1

Cancel the common factor of $3$ .

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Step 11.2.2.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Step 11.2.2.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{3}$

Step 11.2.3

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 11.2.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 11.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 11.2.4

The final solution is all the values that make $(3 x - 5) (x - 1) = 0$ true.

$x = \frac{5}{3}, 1$

Step 11.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 1) \cup (1, \frac{5}{3}) \cup (\frac{5}{3}, \infty)$

Step 12

Use each root to create test intervals.

$x < 1$

$1 < x < \frac{5}{3}$

$\frac{5}{3} < x < 2$

$2 < x < 5$

$x > 5$

Step 13

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 13.1

Test a value on the interval $x < 1$ to see if it makes the inequality true.

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Step 13.1.1

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 13.1.2

Replace $x$ with $0$ in the original inequality.

$\frac{0}{3 (0) - 5} \leq \frac{2}{(0) - 1}$

Step 13.1.3

The left side $0$ is greater than the right side $- 2$ , which means that the given statement is false.

False

Step 13.2

Test a value on the interval $1 < x < \frac{5}{3}$ to see if it makes the inequality true.

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Step 13.2.1

Choose a value on the interval $1 < x < \frac{5}{3}$ and see if this value makes the original inequality true.

$x = 1.33$

Step 13.2.2

Replace $x$ with $1.33$ in the original inequality.

$\frac{1.33}{3 (1.33) - 5} \leq \frac{2}{(1.33) - 1}$

Step 13.2.3

The left side $- 1.\overline{3168}$ is less than the right side $6.\overline{06}$ , which means that the given statement is always true.

True

Step 13.3

Test a value on the interval $\frac{5}{3} < x < 2$ to see if it makes the inequality true.

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Step 13.3.1

Choose a value on the interval $\frac{5}{3} < x < 2$ and see if this value makes the original inequality true.

$x = 1.83$

Step 13.3.2

Replace $x$ with $1.83$ in the original inequality.

$\frac{1.83}{3 (1.83) - 5} \leq \frac{2}{(1.83) - 1}$

Step 13.3.3

The left side $3.73469387$ is greater than the right side $2.40963855$ , which means that the given statement is false.

False

Step 13.4

Test a value on the interval $2 < x < 5$ to see if it makes the inequality true.

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Step 13.4.1

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Step 13.4.2

Replace $x$ with $4$ in the original inequality.

$\frac{4}{3 (4) - 5} \leq \frac{2}{(4) - 1}$

Step 13.4.3

The left side $0.\overline{571428}$ is less than the right side $0.\overline{6}$ , which means that the given statement is always true.

True

Step 13.5

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 13.5.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 13.5.2

Replace $x$ with $8$ in the original inequality.

$\frac{8}{3 (8) - 5} \leq \frac{2}{(8) - 1}$

Step 13.5.3

The left side $0.42105263$ is greater than the right side $0.\overline{285714}$ , which means that the given statement is false.

False

Step 13.6

Compare the intervals to determine which ones satisfy the original inequality.

$x < 1$ False

$1 < x < \frac{5}{3}$ True

$\frac{5}{3} < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

$x < 1$ False

$1 < x < \frac{5}{3}$ True

$\frac{5}{3} < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

Step 14

The solution consists of all of the true intervals.

$1 < x < \frac{5}{3}$ or $2 \leq x \leq 5$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$1 < x < \frac{5}{3} or 2 \leq x \leq 5$

Interval Notation:

$(1, \frac{5}{3}) \cup [2, 5]$

Step 16