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Equations · Inequalities and systems of inequalities
Inequalities
What an inequality is and how to write the solution The map of the topic: six forms of inequality Absolute value and roots: two advances on 4th ESO Systems and several unknowns Summary
An equation usually has a finite number of solutions. An inequality, on the other hand, almost always has infinitely many: a whole stretch of the number line. In 4th ESO you already solved simple inequalities; in this topic you will systematise the method and extend it to cases that were only hinted at in 4th ESO: rational inequalities, inequalities with absolute value, inequalities with roots and systems with two unknowns. The final goal is that, faced with any inequality, you can identify its type and apply the right strategy without hesitation.
Before you start, make sure you know: Inequalities (4th ESO)
What an inequality is and how to write the solution
A comparison between two algebraic expressions, using $<$, $>$, $\leq$ or $\geq$ instead of $=$. To solve it is to find all the values of the unknown that make it true: this set is called the solution set .
The solution set is almost never a single isolated number: it is a stretch of the real number line, which is written as an interval .
Representation on the real number line of the open interval (-1, +infinity), with an open circle at the value -1 and a solid half-line to the right.
x > −1
−1 is not included
−3 −2 0 1 2
−1
(−1, +∞)
Confusing "less than" with "less than or equal to" changes the result: $x<3$ excludes 3, $x\leq3$ includes it. On the number line you can see it in whether the circle at the endpoint is open or filled. Always check this at the end.
If you multiply or divide both sides of an inequality by a negative number, the direction of the inequality reverses (for example, from $<$ to $>$). Forgetting to reverse the sign when isolating the unknown is one of the most common mistakes.
Marcos solves an inequality and arrives at $x \geq 3$. As the solution he writes the interval $[3, +\infty]$. What detail has he overlooked?
Show answer Infinity is never a number that can be reached or included, so it never takes a square bracket. The correct notation is $[3, +\infty)$.
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Solve the following system of linear inequalities in one variable: $2x - 3 < 5$ $4 - x \leq 2$ Solve the following system of linear inequalities in one variable:
$2x - 3 < 5$
$4 - x \leq 2$
Solve
The map of the topic: six forms of inequality
Each section of this topic tackles a different type. Seeing them all together before going into detail helps you recognise which one you are facing.
Linear ($ax+b
Quadratic ($ax^2+bx+c>0$): roots + sign study by intervals.
Rational ($\frac{P(x)}{Q(x)}\geq0$): sign of the numerator and of the denominator, domain excluding where the denominator is zero.
With absolute value ($|x-a|\leq k$): translated into one or two inequalities without absolute value.
With roots ($\sqrt{P(x)}
Systems with two unknowns : the solution is no longer an interval of the line, but a region of the plane.
Representation on the real number line of different types of intervals: open, closed, half-open and infinite, with the endpoints marked by filled or open circles.
(x − 2)(x − 3) ≥ 0
2
3
+ − +
solution: (−∞,2] ∪ [3,+∞)
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Lucía wants to solve the rational inequality $\frac{x-1}{x+3} > 0$ by multiplying both sides directly by $x+3$ to get $x-1 > 0$. What mistake does she make?
Show answer The sign of $x+3$ varies depending on the value of $x$: if $x+3 < 0$, the direction of the inequality would reverse, so you cannot multiply by an expression containing the unknown without first knowing its sign.
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Solve the following linear inequality, expressing the solution in interval notation: $\frac{3x - 1}{2} - \frac{x + 4}{3} \leq \frac{5x}{6} + 1$ Solve the following linear inequality, expressing the solution in interval notation:
$\frac{3x - 1}{2} - \frac{x + 4}{3} \leq \frac{5x}{6} + 1$
Solve
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Find the solution set of the following quadratic inequality: $-2x^2 + 8x - 6 > 0$ Find the solution set of the following quadratic inequality:
$-2x^2 + 8x - 6 > 0$
Solve
Absolute value and roots: two advances on 4th ESO
In 4th ESO you saw the simplest case of absolute value. Now you will solve inequalities with absolute value applied to quadratic or rational expressions, and bring in roots, which require a new precaution: before squaring you must fix the domain (what is inside the root cannot be negative) and, if one side has no root, check its sign.
Comparison of three real number lines illustrating the domain of the root (x greater than or equal to -2), the requirement that the right-hand side be non-negative (x greater than or equal to 0) and their final intersection (x greater than or equal to 0).
√(x + 2) ≤ x
domain: x ≥ −2
also x ≥ 0
both at once: x ≥ 0
−2 0
Squaring both sides only preserves the order if both are positive ($2 < 3 \implies 4 < 9$). If there are negative terms, the inequality can reverse or produce false solutions (for example, $-5 < 2$, but $(-5)^2 = 25 > 4$).
Look at the inequality $\sqrt{2x - 1} \leq -3$. Do you need to square to solve it?
Show answer No, it has no solution. The square root (the principal root) is by definition greater than or equal to zero, so it can never be less than or equal to a negative number.
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Solve the following modulus inequality and represent the solution on the number line: $|2x - 5| < 3$ Solve the following modulus inequality and represent the solution on the number line:
$|2x - 5| < 3$
Solve
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Determine the interval of values of $x$ that satisfy the following radical inequality, taking into account the domain of the root: $\sqrt{2x - 4} < 4$ Determine the interval of values of $x$ that satisfy the following radical inequality, taking into account the domain of the root:
$\sqrt{2x - 4} < 4$
Solve
Systems and several unknowns
Three real number lines showing the solution of x greater than -3, of x less than or equal to 4 and their intersection, the interval (-3, 4], with an open circle at -3 and a filled circle at 4.
A: x > −3
B: x ≤ 4
A ∩ B
(−3, 4]
With two unknowns the logic is the same, but the solution no longer fits on a line: it is a region of the plane . This is the subject of systems of linear inequalities with two unknowns, which you will see in the last section.
Why can't the solution of a system of inequalities with two unknowns be expressed as an interval of the real line?
Show answer Because each solution is an ordered pair $(x,y)$ representing a point of the Cartesian plane, so the solution set is a region of the plane (a half-plane or a polygonal region) and not a segment of the real line.
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For a given chemical compound to remain stable, its temperature $T$ (in degrees Celsius) must simultaneously satisfy two conditions: twice the temperature decreased by 5… For a given chemical compound to remain stable, its temperature $T$ (in degrees Celsius) must simultaneously satisfy two conditions: twice the temperature decreased by 5 degrees must not exceed 35 degrees, and one third of the temperature increased by 10 degrees must be strictly greater than 15 degrees. Set up a system of inequalities and find the temperature range in which the compound remains stable.
Solve
Summary
An inequality has as its solution a set of values, which is written as an interval: $(a,b)$, $[a,b]$, with $\infty$ always in round brackets.
This topic covers six types: linear, quadratic, rational, with absolute value, with roots and systems with two unknowns.
Rational inequalities and inequalities with roots require you to watch the domain (denominator $\neq0$, radicand $\geq0$) before operating.
A system is solved by taking the intersection of the solutions of each inequality.
← Previous Non-linear systemsNext → Linear inequalities in one variable