Pre-Algebra Examples

$2 | 2 x - 3 | < | x - 10 |$

Step 1

Replace $<$ with $=$ in $2 | 2 x - 3 | < | x - 10 |$ .

$2 | 2 x - 3 | = | x - 10 |$

Step 2

Rewrite the absolute value equation as four equations without absolute value bars.

$2 (2 x - 3) = x - 10$

$2 (2 x - 3) = - (x - 10)$

$- 2 (2 x - 3) = x - 10$

$- 2 (2 x - 3) = - (x - 10)$

Step 3

After simplifying, there are only two unique equations to be solved.

$2 (2 x - 3) = x - 10$

$2 (2 x - 3) = - (x - 10)$

Step 4

Solve $2 (2 x - 3) = x - 10$ for $x$ .

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Step 4.1

Simplify $2 (2 x - 3)$ .

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Step 4.1.1

Rewrite.

$0 + 0 + 2 (2 x - 3) = x - 10$

Step 4.1.2

Simplify by adding zeros.

$2 (2 x - 3) = x - 10$

Step 4.1.3

Apply the distributive property.

$2 (2 x) + 2 \cdot - 3 = x - 10$

Step 4.1.4

Multiply.

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Step 4.1.4.1

Multiply $2$ by $2$ .

$4 x + 2 \cdot - 3 = x - 10$

Step 4.1.4.2

Multiply $2$ by $- 3$ .

$4 x - 6 = x - 10$

Step 4.2

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

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Step 4.2.1

Subtract $x$ from both sides of the equation.

$4 x - 6 - x = - 10$

Step 4.2.2

Subtract $x$ from $4 x$ .

$3 x - 6 = - 10$

Step 4.3

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

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Step 4.3.1

Add $6$ to both sides of the equation.

$3 x = - 10 + 6$

Step 4.3.2

Add $- 10$ and $6$ .

$3 x = - 4$

Step 4.4

Divide each term in $3 x = - 4$ by $3$ and simplify.

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Step 4.4.1

Divide each term in $3 x = - 4$ by $3$ .

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 4.4.2

Simplify the left side.

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Step 4.4.2.1

Cancel the common factor of $3$ .

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Step 4.4.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 4}{3}$

Step 4.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{3}$

Step 4.4.3

Simplify the right side.

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Step 4.4.3.1

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

Step 5

Solve $2 (2 x - 3) = - (x - 10)$ for $x$ .

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Step 5.1

Simplify $2 (2 x - 3)$ .

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Step 5.1.1

Rewrite.

$0 + 0 + 2 (2 x - 3) = - (x - 10)$

Step 5.1.2

Simplify by adding zeros.

$2 (2 x - 3) = - (x - 10)$

Step 5.1.3

Apply the distributive property.

$2 (2 x) + 2 \cdot - 3 = - (x - 10)$

Step 5.1.4

Multiply.

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Step 5.1.4.1

Multiply $2$ by $2$ .

$4 x + 2 \cdot - 3 = - (x - 10)$

Step 5.1.4.2

Multiply $2$ by $- 3$ .

$4 x - 6 = - (x - 10)$

Step 5.2

Simplify $- (x - 10)$ .

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Step 5.2.1

Apply the distributive property.

$4 x - 6 = - x - - 10$

Step 5.2.2

Multiply $- 1$ by $- 10$ .

$4 x - 6 = - x + 10$

Step 5.3

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

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Step 5.3.1

Add $x$ to both sides of the equation.

$4 x - 6 + x = 10$

Step 5.3.2

Add $4 x$ and $x$ .

$5 x - 6 = 10$

Step 5.4

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

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Step 5.4.1

Add $6$ to both sides of the equation.

$5 x = 10 + 6$

Step 5.4.2

Add $10$ and $6$ .

$5 x = 16$

Step 5.5

Divide each term in $5 x = 16$ by $5$ and simplify.

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Step 5.5.1

Divide each term in $5 x = 16$ by $5$ .

$\frac{5 x}{5} = \frac{16}{5}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $5$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{16}{5}$

Step 5.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{16}{5}$

Step 6

List all of the solutions.

$x = - \frac{4}{3}, \frac{16}{5}$

Step 7

Use each root to create test intervals.

$x < - \frac{4}{3}$

$- \frac{4}{3} < x < \frac{16}{5}$

$x > \frac{16}{5}$

Step 8

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 8.1

Test a value on the interval $x < - \frac{4}{3}$ to see if it makes the inequality true.

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Step 8.1.1

Choose a value on the interval $x < - \frac{4}{3}$ and see if this value makes the original inequality true.

$x = - 4$

Step 8.1.2

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

$2 | 2 (- 4) - 3 | < | (- 4) - 10 |$

Step 8.1.3

The left side $22$ is not less than the right side $14$ , which means that the given statement is false.

False

Step 8.2

Test a value on the interval $- \frac{4}{3} < x < \frac{16}{5}$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $- \frac{4}{3} < x < \frac{16}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 8.2.2

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

$2 | 2 (0) - 3 | < | (0) - 10 |$

Step 8.2.3

The left side $6$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 8.3

Test a value on the interval $x > \frac{16}{5}$ to see if it makes the inequality true.

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Step 8.3.1

Choose a value on the interval $x > \frac{16}{5}$ and see if this value makes the original inequality true.

$x = 6$

Step 8.3.2

Replace $x$ with $6$ in the original inequality.

$2 | 2 (6) - 3 | < | (6) - 10 |$

Step 8.3.3

The left side $18$ is not less than the right side $4$ , which means that the given statement is false.

False

Step 8.4

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

$x < - \frac{4}{3}$ False

$- \frac{4}{3} < x < \frac{16}{5}$ True

$x > \frac{16}{5}$ False

$x < - \frac{4}{3}$ False

$- \frac{4}{3} < x < \frac{16}{5}$ True

$x > \frac{16}{5}$ False

Step 9

The solution consists of all of the true intervals.

$- \frac{4}{3} < x < \frac{16}{5}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$- \frac{4}{3} < x < \frac{16}{5}$

Interval Notation:

$(- \frac{4}{3}, \frac{16}{5})$

Step 11