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Equations · Inequalities and systems of inequalities

Inequalities with absolute value

The absolute value $|x-a|$ measures the distance between $x$ and $a$ on the real number line, regardless of which side it lies on. This geometric idea is the key to the whole section: it turns any inequality with an absolute value into one (or two) ordinary inequalities.

Before you start, make sure you know: Intervals, rays and neighbourhoods (4th ESO)
Before you start, make sure you know: Inequalities with one unknown (4th ESO)

Absolute value as distance

Geometric representation on the real number line of absolute value as the distance between the points 3 and x. |x − 3| = distance between x and 3 distance 3x
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Solve the inequality $|x - 5| < 3$ and express the solution as an interval.

Solve the inequality $|x - 5| < 3$ and express the solution as an interval.

Solve
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Find all real numbers $x$ that satisfy the inequality $|2x + 4| \geq 8$.

Find all real numbers $x$ that satisfy the inequality $|2x + 4| \geq 8$.

Solve

Cases with no solution or with the whole line as the solution

Since $|x|\geq0$ always, there are inequalities with absolute value that need no calculation at all: they are solved just by reasoning about the sign.

Representation on the real number line of the open interval between 2 and 8, centred at 5 with radius 3. |2x − 5| < −1no solution (negative distance, impossible) |x + 4| ≥ −3always true whenever x + 4 exists |x + 4| ≤ 0only x = −4
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Justify whether the following inequalities have a solution in the set of real numbers and, if so, find it: a) $|x + 4| \leq 0$ b) $|2x - 5| < -1$

Justify whether the following inequalities have a solution in the set of real numbers and, if so, find it:

  1. a) $|x + 4| \leq 0$
  2. b) $|2x - 5| < -1$
Solve

Tolerances and margins of error

The most common real-life context for absolute value is a tolerance: a measurement that must be close to an ideal value, with a maximum margin of error.

Symmetric tolerance interval on the real number line centred at 12,5 with a margin of error of 0,15. |L − 12,5| ≤ 0,15 0,150,15 12,3512,512,65
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In an electronic components factory, the ideal length of a chip is $12,5$ mm. For quality control reasons, a maximum margin of error of $0,15$ mm is accepted. Write a…

In an electronic components factory, the ideal length of a chip is $12,5$ mm. For quality control reasons, a maximum margin of error of $0,15$ mm is accepted. Write a modulus inequality that models this situation for the length $L$ of the chip and solve it to find the interval of acceptable lengths.

Solve
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The pH of a swimming pool must be maintained at an optimal level of $7,4$. For the water to be suitable for swimming, the absolute difference between the current pH and…

The pH of a swimming pool must be maintained at an optimal level of $7,4$. For the water to be suitable for swimming, the absolute difference between the current pH and the optimal pH must not exceed $5\%$ of the optimal value. Set up the corresponding modulus inequality and determine the range of permitted pH levels.

Solve

Absolute value applied to quadratic or rational expressions, or with another absolute value

When there is a quadratic or rational expression inside the absolute value, the method does not change: it is translated into a double inequality or into two sections, and then that inequality (which may now be quadratic or rational) is solved using the techniques from the previous sections. If there is an absolute value on both sides, we square both sides or split into cases.

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Solve the quadratic modulus inequality $|x^2 - 5| < 4$.

Solve the quadratic modulus inequality $|x^2 - 5| < 4$.

Solve
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Solve the rational modulus inequality $\frac{2}{|x - 3|} \geq 1$.

Solve the rational modulus inequality $\frac{2}{|x - 3|} \geq 1$.

Solve
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Find the solution set of the inequality $|x - 1| + |x - 3| \leq 6$. The figure shows the graph of $f(x) = |x-1| + |x-3|$ (in blue) and the line $y = 6$ (in green). You…

Find the solution set of the inequality $|x - 1| + |x - 3| \leq 6$.

The figure shows the graph of $f(x) = |x-1| + |x-3|$ (in blue) and the line $y = 6$ (in green). You can use it to check the result you obtain algebraically.

Graph of f(x) = |x−1| + |x−3|, V-shaped with a flat bottom, and the horizontal line y = 6−2−1012345672468xyf(x) = |x − 1| + |x − 3|y = 6
Solve

Summary

See all the exercises in this section