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  1. Home
  2. 12th grade
  3. Precalculus
  4. Exercise : Solve inequalities involving absolute values with calculations

Solve inequalities involving absolute values with calculations Precalculus

Solve the following inequalities.

\left| 5x-1 \right| \lt 3

Consider the following inequality:

\left| 5x-1\right| \lt 3

It is equivalent to:

-3 \lt 5x-1 \lt 3

Adding 1 to all sides gives:

-2 \lt 5x \lt 4

Dividing all sides by 5 gives:

-\dfrac{2}{5} \lt x \lt \dfrac{4}{5}

x is a solution if and only if -\dfrac{2}{5} \lt x \lt \dfrac{4}{5}.

\left| 2x+7 \right| \leqslant9

Consider the following inequality:

\left| 2x+7 \right| \leqslant9

It is equivalent to:

-9 \leqslant 2x+7 \leqslant 9

Subtracting 7 from all sides gives:

-16 \leqslant 2x \leqslant 2

Dividing all sides by 2 gives:

-8\leqslant x\leqslant1

x is a solution if and only if -8\leqslant x\leqslant1.

\left| x+3 \right| \gt 10

Consider the following inequality:

\left| x+3 \right| \gt 10

It is equivalent to:

x+3 \gt 10 or x+3 \lt -10

Subtracting 3 from all sides in the two equations gives:

x \gt 7 or x \lt -13

x is a solution if and only if x \lt -13 or x \gt 7.

\left| \dfrac{2x-3}{7}\right| \lt 5

Consider the following inequality:

\left| \dfrac{2x-3}{7}\right| \lt 5

It is equivalent to:

-5 \lt \dfrac{2x-3}{7} \lt 5

Multiplying all sides by 7 gives:

-35 \lt 2x-3 \lt 35

Adding 3 to all sides gives:

-32 \lt 2x \lt 38

Dividing all sides by 2 gives:

-16 \lt x \lt 19

x is a solution if and only if -16 \lt x \lt 19.

\left| \dfrac{x}{3}-4\right| \geqslant1

Consider the following inequality:

\left| \dfrac{x}{3}-4\right| \geqslant1

It is equivalent to:

\dfrac{x}{3}-4\geq 1 or \dfrac{x}{3} - 4 \leq -1

Adding 4 to all sides in the two equations gives:

\dfrac{x}{3}\geq5 or \dfrac{x}{3}\leq3

Multiplying all sides by 3 in the two equations gives:

x\geq15 or x\leq9

x is a solution if and only if x\leqslant9 or x\geqslant15.

\left| \dfrac{2x-5}{3}+6 \right| \lt 1

Consider the following inequality:

\left| \dfrac{2x-5}{3}+6 \right| \lt 1

It is equivalent to:

-1 \lt \dfrac{2x-5}{3}+6 \lt 1

Subtracting 6 from all sides gives:

-7 \lt \dfrac{2x-5}{3} \lt -5

Multiplying all sides by 3 gives:

-21 \lt 2x-5 \lt -15

Adding 5 to all sides gives:

-16 \lt 2x \lt -10

Dividing all sides by 2 gives:

-8 \lt x \lt -5

x is a solution if and only if -8 \lt x \lt -5.

\left| -x+5 \right| \lt 3

Consider the following inequality:

\left| -x+5 \right| \lt 3

It is equivalent to:

-3 \lt -x+5 \lt 3

Subtracting 5 from all sides gives:

-8 \lt -x \lt -2

Multiplying all sides by -1 gives:

8 \gt x \gt 2

Which is equivalent to:

2 \lt x \lt 8

x is a solution if and only if 2 \lt x \lt 8.

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See also
  • Course : Absolute value function
  • Exercise : Calculate expressions involving absolute values
  • Exercise : Graph functions involving absolute values
  • Exercise : Solve equations involving absolute values with calculations
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