Linear inequalities and changing the subject
Orla's mobile plan costs €12 a month plus 5c per minute of calls. She wants to spend at most €20 in a month. How many minutes can she talk? An equation gives one answer, but here the answer is a whole range of values. That is what an inequality describes. We will come back to Orla's phone at the end of the section on problems.
In the second half of the page you will learn the other big algebra skill: rearranging a formula so that a different letter is on its own.
Inequality symbols and the number line
An inequality compares two quantities that are not necessarily equal.
The solution of an equation such as $x + 2 = 5$ is a single number. The solution of $x + 2 > 5$ is every number bigger than 3, so we show it on a number line.
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An open circle ○ means the number is not included ($<$ or $>$).
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A closed circle ● means the number is included ($\le$ or $\ge$).
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A thick line and arrow show all the other values that work.
The symbol $\mathbb{R}$ means we allow every real number (including fractions and decimals). Other sets you will meet: $\mathbb{Z}$ (integers: $\dots, -2, -1, 0, 1, 2, \dots$) and $\mathbb{N}$ (natural numbers: $1, 2, 3, \dots$).
Solving: do the same to both sides
Solving a linear inequality works just like solving a linear equation. Whatever you add, subtract, multiply or divide on one side, you do on the other.
The balance above shows the equation $2x + 1 = 9$. For an inequality the steps are exactly the same, except that the answer is a range of values rather than one.
The one new rule
Why? Start with a true statement: $2 < 5$. Multiply both sides by $-1$: you get $-2$ and $-5$. On the number line $-5$ is to the left of $-2$, so $-2 > -5$. The order flipped.
Solve each inequality for $x \in \mathbb{R}$. a) $x + 5 > 9$ b) $3x \le 18$ c) $2x - 1 < 7$ d) $-x \ge 4$
Solve each inequality for $x \in \mathbb{R}$.
- a) $x + 5 > 9$
- b) $3x \le 18$
- c) $2x - 1 < 7$
- d) $-x \ge 4$
Variables on both sides, brackets and double inequalities
When $x$ appears on both sides, collect the $x$ terms on one side and the numbers on the other, just as for equations.
If there are fractions, multiply every term by the common denominator first (a positive number, so the sign stays).
Solutions in $\mathbb{N}$ or $\mathbb{Z}$
The set matters. If $x \in \mathbb{R}$ the solution is a continuous line. If $x \in \mathbb{Z}$ or $x \in \mathbb{N}$ you list separate whole numbers and show them as dots.
Double inequalities
A double inequality such as $-2 < x \le 4$ means "$x$ is greater than $-2$ and less than or equal to 4". To solve one, do the same operation on all three parts.
Solve each inequality for $x \in \mathbb{N}$ and list the solutions. a) $x - 2 < 3$ b) $2x \le 9$ c) $x + 1 \ge 4$ and $x < 7$
Solve each inequality for $x \in \mathbb{N}$ and list the solutions.
- a) $x - 2 < 3$
- b) $2x \le 9$
- c) $x + 1 \ge 4$ and $x < 7$
Solve each inequality for $x \in \mathbb{R}$ and graph the solution on a number line. a) $5x - 3 \le 2x + 9$ b) $\frac{x}{3} + 2 > 4$ c) $4 - 3x < 13$
Solve each inequality for $x \in \mathbb{R}$ and graph the solution on a number line.
- a) $5x - 3 \le 2x + 9$
- b) $\frac{x}{3} + 2 > 4$
- c) $4 - 3x < 13$
Solve $-3 < 2x + 1 \le 9$ for $x \in \mathbb{Z}$, list the integer solutions and show them on a number line.
Solve $-3 < 2x + 1 \le 9$ for $x \in \mathbb{Z}$, list the integer solutions and show them on a number line.
Inequalities in context
To solve a worded problem, follow the same plan as for equations:
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Define the unknown (say what $x$ stands for, with units).
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Translate the words into an inequality.
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Solve it.
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Interpret the answer in the context and check it is sensible.
Useful phrases: "at most" means $\le$; "at least" means $\ge$; "fewer than" means $<$; "more than" means $>$.
Comparing two deals gives an inequality with letters on both sides.
A taxi in Dublin charges a fixed fee of €4 plus €1.20 per kilometre. Ciara has at most €30 to spend on a journey. a) Write an inequality for the number of kilometres $k$…
A taxi in Dublin charges a fixed fee of €4 plus €1.20 per kilometre. Ciara has at most €30 to spend on a journey.
- a) Write an inequality for the number of kilometres $k$ she can travel.
- b) Solve it and state the greatest distance, correct to one decimal place.
- c) A second company charges €2 fixed plus €1.50 per kilometre. Find the values of $k$ for which the first company is cheaper.
Changing the subject of a formula
Higher Level only.
In the formula $A = \pi r^2$ the subject is $A$: it stands alone on the left. If you know the area and want the radius, it is easier to rearrange the formula so that $r$ is the subject.
Rearranging a formula is just solving an equation where the "numbers" are letters. The method is the same: do the same thing to both sides, undoing operations in the reverse order of BIDMAS.
Formulae in geometry are a good place to practise.
Make $x$ the subject of each formula. a) $y = x + 7$ b) $y = 4x$ c) $y = 3x - 5$
Make $x$ the subject of each formula.
- a) $y = x + 7$
- b) $y = 4x$
- c) $y = 3x - 5$
Make $r$ the subject of each formula. a) $A = \pi r^2$, where $r > 0$ b) $C = 2\pi r$ c) $V = \frac{1}{3}\pi r^2 h$, where $r > 0$
Make $r$ the subject of each formula.
- a) $A = \pi r^2$, where $r > 0$
- b) $C = 2\pi r$
- c) $V = \frac{1}{3}\pi r^2 h$, where $r > 0$
Harder formulae: squares, roots and fractions
Higher Level only.
Powers and roots
To remove a square, take the square root. To remove a square root, square both sides.
The letter appears twice
If the new subject appears in more than one place, gather all the terms containing it on one side, then factorise it out.
Fractions with reciprocals
The formula $s = ut + \frac{1}{2}at^2$ gives the distance travelled. a) Make $a$ the subject. b) Make $u$ the subject. c) Find $a$ when $s = 100$, $u = 5$ and $t = 8$.
The formula $s = ut + \frac{1}{2}at^2$ gives the distance travelled.
- a) Make $a$ the subject.
- b) Make $u$ the subject.
- c) Find $a$ when $s = 100$, $u = 5$ and $t = 8$.
Make $x$ the subject of each formula, and state any restriction on the variables. a) $y = \frac{2x + 3}{x - 1}$ b) $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ (make $v$…
Make $x$ the subject of each formula, and state any restriction on the variables.
- a) $y = \frac{2x + 3}{x - 1}$
- b) $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ (make $v$ the subject)
- c) $T = 2\pi\sqrt{\frac{l}{g}}$ (make $l$ the subject)
Summary
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An inequality uses $<$, $\le$, $>$, $\ge$. On a number line use an open circle when the end is not included and a closed circle when it is.
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Solve linear inequalities like equations, but reverse the sign when you multiply or divide by a negative number.
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For a double inequality such as $-2 < x \le 4$, do the same operation on all three parts.
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The set matters: for $x \in \mathbb{R}$ give a range, for $x \in \mathbb{Z}$ or $x \in \mathbb{N}$ list the whole numbers.
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In problems, "at most" means $\le$ and "at least" means $\ge$; interpret and check your answer in context.
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(Higher Level) To change the subject, undo operations in reverse order, divide the whole side, factorise if the letter appears twice, and state any restrictions (no division by zero).