International Baccalaureate · MYP 5 Extended

Linear Equations, Inequalities and Formulae

⌚ ~130 min ▢ 14 blocks ✓ 55 exercises ➣ Prerequisites : Expansion and Factorisation, Algebraic Fractions, Linear Equations, Inequalities and Formulae, Sets, Number Sets and Intervals

I Solving Equations with Algebraic Fractions

Definition — Excluded Values

Let $\dfrac{A}{B}$ be an algebraic fraction in the variable $x$. The values of $x$ for which the denominator $B$ is equal to $0$ are the excluded values of the fraction (also called forbidden values): for these values, the fraction has no value. An expression that contains several algebraic fractions has as excluded values those of all its denominators and, in a quotient, the values for which the dividing fraction is $0$.
For example, $\dfrac{x}{x^2 - 9}$ has the excluded values $3$ and $-3$.

In an equation such as $\dfrac{x}{x - 1} - \dfrac{2}{x + 1} = 1$, the unknown appears in denominators. Multiplying both sides by a common denominator turns it into an equation without fractions, which is solved as usual — but an excluded value can be a solution of this new equation without being a solution of the original one.

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Proposition — Clearing the Denominators

Consider an equation that contains algebraic fractions in $x$, and let $D$ be the lowest common denominator of all its fractions. For every $x$ that is not an excluded value, $D \neq 0$, and the equation is equivalent to the equation obtained by multiplying each term of both sides by $D$, which contains no fraction. Hence the solutions of the equation are the solutions of this cleared equation that are not excluded values.

Proof

Let $x$ be a value that is not excluded. Every denominator of the equation is non-zero at $x$, so $D$, a product of their factors, is non-zero too. By the Golden Rules of Solving Equations, multiplying both sides by the non-zero number $D$ gives an equivalent equation; multiplying a side by $D$ multiplies each of its terms by $D$ (distributivity), and in each term the denominator cancels, since it divides $D$. So, among the values that are not excluded, the equation and the cleared equation have the same solutions.
An excluded value is never a solution: at that value, a fraction of the equation has no value, so the equation has no meaning — even if the cleared equation happens to be true there.

Method — Solving an Equation with Algebraic Fractions

Factorise the denominators and write the excluded values.
Multiply each term of both sides by the LCD and simplify each term: no fraction remains.
Solve the cleared equation: a linear equation, or a quadratic equation brought to the form $\ldots = 0$, factorised, and solved with the Null Factor Law (Method “Solving by Factorisation”).
Reject every solution that is an excluded value.
Check each remaining solution in the original equation.

Example

Solve $\dfrac{x}{x - 1} - \dfrac{2}{x + 1} = 1$.

Answer

The excluded values are $1$ and $-1$, and the LCD is $(x - 1)(x + 1)$. Multiplying the three terms by it gives $\dfrac{x}{x - 1} \times (x - 1)(x + 1) = x(x + 1)$, $\dfrac{2}{x + 1} \times (x - 1)(x + 1) = 2(x - 1)$ and $1 \times (x - 1)(x + 1)$. For $x \neq 1$ and $x \neq -1$: $$\begin{aligned}[t] \frac{x}{x - 1} - \frac{2}{x + 1} &= 1 \\ x(x + 1) - 2(x - 1) &= (x - 1)(x + 1) && \text{(multiply each term by the LCD)} \\ x^2 + x - 2x + 2 &= x^2 - 1 && \text{(expand)} \\ -x + 2 &= -1 && \text{(subtract } x^2 \text{ from both sides)} \\ -x &= -3 && \text{(subtract } 2\text{)} \\ x &= 3. \end{aligned}$$ The value $3$ is not excluded. Check: $\dfrac{3}{2} - \dfrac{2}{4} = \dfrac{3}{2} - \dfrac{1}{2} = 1$. The equation has one solution, $x = 3$.

Example

Solve each equation.

$\dfrac{1}{x - 2} + \dfrac{x}{x + 2} = \dfrac{8}{x^2 - 4}$
$\dfrac{2}{x - 1} - \dfrac{1}{x + 1} = \dfrac{4}{x^2 - 1}$

Answer

$x^2 - 4 = (x - 2)(x + 2)$: the excluded values are $2$ and $-2$, and the LCD is $(x - 2)(x + 2)$. For $x \neq 2$ and $x \neq -2$: $$\begin{aligned}[t] \frac{1}{x - 2} + \frac{x}{x + 2} &= \frac{8}{x^2 - 4} \\ (x + 2) + x(x - 2) &= 8 && \text{(multiply each term by the LCD)} \\ x^2 - x + 2 &= 8 && \text{(expand and collect)} \\ x^2 - x - 6 &= 0 && \text{(subtract } 8\text{)} \\ (x - 3)(x + 2) &= 0. && \text{(sum } -1\text{, product } -6\text{)} \end{aligned}$$ By the Null Factor Law, $x = 3$ or $x = -2$. The value $-2$ is excluded: it is rejected. Check $x = 3$: $\dfrac{1}{1} + \dfrac{3}{5} = \dfrac{8}{5}$ and $\dfrac{8}{9 - 4} = \dfrac{8}{5}$. The equation has one solution, $x = 3$.
$x^2 - 1 = (x - 1)(x + 1)$: the excluded values are $1$ and $-1$. For $x \neq 1$ and $x \neq -1$: $$\begin{aligned}[t] \frac{2}{x - 1} - \frac{1}{x + 1} &= \frac{4}{x^2 - 1} \\ 2(x + 1) - (x - 1) &= 4 && \text{(multiply each term by } (x - 1)(x + 1)\text{)} \\ x + 3 &= 4 && \text{(expand and collect)} \\ x &= 1. \end{aligned}$$ The only solution of the cleared equation, $1$, is excluded: it is rejected, and the equation has no solution.

Skills to practice

Counting the Solutions of an Equation
Solving Equations with Algebraic Fractions
Solving Equations and Problems with Algebraic Fractions

II Defining an Inequality

An inequality is a mathematical statement that one expression is greater than or less than another. While an equation typically has one or a few specific solutions, an inequality describes a whole range (or interval) of values. Understanding inequalities is essential for modelling real-world constraints, such as budget limits, speed limits, or acceptable temperature ranges.

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Definition — Inequality

An inequality is a statement comparing two expressions using one of the following symbols:

$<$ (less than)
$>$ (greater than)
$\leq$ (less than or equal to)
$\geq$ (greater than or equal to)

The signs $<$ and $>$ are called strict, the signs $\leq$ and $\geq$ non-strict. To solve an inequality means to find all values of the variable that make the statement true.

Example

Is $2$ a solution of the inequality $3x - 1 < 5$? Of the inequality $3x - 1 \leq 5$?
For $x = 2$: $3 \times 2 - 1 = 5$. The statement $5 < 5$ is false, so $2$ is not a solution of $3x - 1 < 5$: with a strict sign, equality is not enough. The statement $5 \leq 5$ is true, so $2$ is a solution of $3x - 1 \leq 5$: with a non-strict sign, equality is enough.

Skills to practice

Defining an Inequality

III Comparing Two Numbers

Method — Comparing Two Numbers by Their Difference or by Their Ratio

To compare two numbers $a$ and $b$ without placing them on a number line, use one of two tools.

By the difference (additive comparison): compute $a - b$ and find its sign. $$a - b > 0 \iff a > b, \qquad a - b = 0 \iff a = b, \qquad a - b < 0 \iff a < b.$$ The difference says how much more: $a = b + (a - b)$.
By the ratio (multiplicative comparison), when $a$ and $b$ are strictly positive: compute $\dfrac{a}{b}$ and compare it to $1$. $$\frac{a}{b} > 1 \iff a > b, \qquad \frac{a}{b} = 1 \iff a = b, \qquad \frac{a}{b} < 1 \iff a < b.$$ The ratio says how many times as much: $a = b \times \dfrac{a}{b}$.

In this lesson the ratio is used only for strictly positive numbers: with negative numbers it can mislead ($\dfrac{-6}{-3} = 2 > 1$, yet $-6 < -3$). The difference always decides.

Example

Compare $\dfrac{3}{7}$ and $\dfrac{2}{5}$ by their difference.
Compare $2\sqrt{2}$ and $3$ by their ratio.
A town has $120\,000$ inhabitants, another has $80\,000$. Compare the two populations additively, then multiplicatively.

Answer

$\dfrac{3}{7} - \dfrac{2}{5} = \dfrac{15}{35} - \dfrac{14}{35} = \dfrac{1}{35} > 0$, so $\dfrac{3}{7} > \dfrac{2}{5}$: the first fraction exceeds the second by $\dfrac{1}{35}$.
Both numbers are strictly positive. $\dfrac{2\sqrt{2}}{3} = \dfrac{\sqrt{4} \times \sqrt{2}}{\sqrt{9}} = \dfrac{\sqrt{8}}{\sqrt{9}}$, and $8 < 9$ gives $\sqrt{8} < \sqrt{9}$ (the square root function is increasing), so the ratio is less than $1$: $2\sqrt{2} < 3$.
Additively: $120\,000 - 80\,000 = 40\,000 > 0$; the first town has $40\,000$ inhabitants more. Multiplicatively: $\dfrac{120\,000}{80\,000} = 1.5 > 1$; the first town has $1.5$ times as many inhabitants, that is $50\pourcent$ more. The two comparisons describe the same situation: an additive change of $+40\,000$, a multiplicative change by the factor $1.5$.

Skills to practice

Choosing the Right Comparison
Comparing Two Numbers by a Computation

IV Solving Double Inequalities

Method — Rules for Solving Inequalities

Solving inequalities is almost identical to solving equations. The same operation can be performed on both sides to isolate the unknown, but with one critical exception.

Adding any number to both sides, or subtracting it from both sides, preserves the inequality.
Multiplying or dividing both sides by a positive number preserves the inequality.
Multiplying or dividing both sides by a negative number reverses the direction of the inequality symbol. For example, $<$ becomes $>$.

Definition — Double Inequality

A double inequality such as $a < E \le b$ states two conditions at once: $a < E$ and $E \le b$. A value is a solution when it satisfies both conditions.

Method — Solving a Double Inequality

Operate on all three parts at once, exactly as on the two sides of a simple inequality:

adding or subtracting a number preserves both symbols;
multiplying or dividing by a positive number preserves both symbols;
multiplying or dividing by a negative number reverses both symbols — and the two bounds swap roles.

Adding Two Double Inequalities

Two double inequalities in the same direction can be added part by part: if $a < x < b$ and $c < y < d$, then $$a + c < x + y < b + d.$$ Indeed, $a < x$ gives $a + c < x + c$, and $c < y$ gives $x + c < x + y$; in the same way, $x + y < x + d < b + d$. The same holds with the non-strict symbols $\leq$.
Double inequalities are never subtracted part by part. To find bounds for $x - y$, first find bounds for $-y$ by multiplying $c < y < d$ by $-1$, which reverses both symbols: $-d < -y < -c$. Then add: $a - d < x - y < b - c$.

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Example

Solve $-3 \le 2x + 1 < 7$. $$\begin{aligned}[t] -3 &\le 2x + 1 < 7 \\ -4 &\le 2x < 6 && \text{(subtract 1 from all three parts)} \\ -2 &\le x < 3. && \text{(divide all three parts by 2)} \end{aligned}$$ The solution set is the interval $[-2,3)$.

Example

Solve $1 < 3 - 2x \le 5$. $$\begin{aligned}[t] 1 &< 3 - 2x \le 5 \\ -2 &< -2x \le 2 && \text{(subtract 3 from all three parts)} \\ 1 &> x \ge -1 && \text{(divide by $-2$: both symbols reverse)} \\ -1 &\le x < 1. && \text{(rewrite in increasing order)} \end{aligned}$$ The solution set is $[-1,1)$. Note that both inequality symbols reversed, and the strict symbol followed the bound it belonged to.

Skills to practice

Solving Double Inequalities

V Making a Repeated Variable the Subject

Method — Making a Repeated Variable the Subject

When the new subject appears in more than one term, inverse operations alone cannot isolate it. If, once the fractions are cleared and the brackets expanded, the new subject appears only to the first power (never squared, never under a root), proceed in four steps:

clear the fractions: multiply both sides by the denominators, stating that each denominator containing a variable is not zero;
expand, then collect all the terms that contain the new subject on one side and all the other terms on the other side;
factorise the new subject out of its terms;
divide both sides by the bracket, which must not be zero.

When the subject appears squared after these steps — as in $y = x^2 + x$, or in $y = \dfrac{x}{x - 1} + \dfrac{x}{x + 1}$, which gives $y(x^2 - 1) = 2x^2$ once the fractions are cleared — factorising leaves it inside the bracket: this Method does not apply.

Example

With simple interest, the amount after $t$ years of a principal $P$ invested at the annual rate $r$ is $A = P + Prt$, with $r > 0$ and $t > 0$. Make $P$ the subject.
Make $x$ the subject of $y = \dfrac{x + 3}{x - 2}$, where $x \neq 2$.

Answer

$P$ appears in two terms. There is no fraction to clear, so we factorise $P$ at once. $$\begin{aligned} A &= P + Prt \\ A &= P(1 + rt) && \text{(factorising $P$)} \\ P &= \dfrac{A}{1 + rt} && \text{(dividing both sides by $1 + rt$)} \end{aligned}$$ The division is allowed: $r > 0$ and $t > 0$, so $1 + rt > 0$.
$x$ appears in the numerator and in the denominator. $$\begin{aligned} y &= \dfrac{x + 3}{x - 2} \\ y(x - 2) &= x + 3 && \text{(multiplying both sides by $x - 2$)} \\ xy - 2y &= x + 3 && \text{(expanding)} \\ xy - x &= 2y + 3 && \text{(collecting the terms in $x$ on the left)} \\ x(y - 1) &= 2y + 3 && \text{(factorising $x$)} \\ x &= \dfrac{2y + 3}{y - 1} && \text{(dividing by $y - 1$, for $y \neq 1$)} \end{aligned}$$ Check with $x = 3$: $y = \dfrac{3 + 3}{3 - 2} = 6$, and the new formula gives back $x = \dfrac{2 \times 6 + 3}{6 - 1} = \dfrac{15}{5} = 3$. The value $y = 1$ is never reached: $x(1 - 1) = 2 + 3$ would read $0 = 5$.

Skills to practice

Making a Repeated Variable the Subject
Rearranging Formulae with a Repeated Variable

VI Predicting Formulae from Differences

A table of values often comes from a situation: a pattern of tiles, a count, a measurement. When its inputs are equally spaced, the differences between consecutive outputs reveal which kind of rule can produce it: a linear rule $y = mx + c$ or a quadratic rule $y = ax^2 + bx + c$.

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Definition — First and Second Differences

In a table whose inputs increase by the same step:

the first differences are the differences between consecutive outputs, $y_2 - y_1$, $y_3 - y_2$, and so on;
the second differences are the differences between consecutive first differences.

Proposition — Differences of Linear and Quadratic Rules

Let the inputs $x$, $x + h$, $x + 2h$, … be equally spaced, with a step $h > 0$.

If $y = mx + c$, every first difference equals $mh$.
If $y = ax^2 + bx + c$ with $a \neq 0$, the first differences are not all equal, and every second difference equals $2ah^2$. For a step $h = 1$, every second difference equals $2a$.

Proof

Linear rule: the first difference that starts at $x$ is $$\big(m(x + h) + c\big) - (mx + c) = mx + mh + c - mx - c = mh,$$ the same number whatever $x$ is.
Quadratic rule: the first difference that starts at $x$ is $$\begin{aligned} d(x) &= \big(a(x + h)^2 + b(x + h) + c\big) - (ax^2 + bx + c) \\ &= ax^2 + 2ahx + ah^2 + bx + bh + c - ax^2 - bx - c && \text{(expanding $(x + h)^2$)} \\ &= 2ahx + ah^2 + bh. && \text{(collecting like terms)} \end{aligned}$$ Since $a \neq 0$ and $h \neq 0$, $d(x)$ changes with $x$: the first differences are not all equal. The second difference that starts at $x$ is $$\begin{aligned} d(x + h) - d(x) &= \big(2ah(x + h) + ah^2 + bh\big) - (2ahx + ah^2 + bh) \\ &= 2ah^2, && \text{(collecting like terms)} \end{aligned}$$ the same number whatever $x$ is; for $h = 1$ it equals $2a$.

Method — Predicting a Rule from a Table of Values

Check that the inputs increase by the same step $h$. Otherwise the differences cannot be used directly.
Compute the first differences. If they are all equal to $d$, predict a linear rule $y = mx + c$ with $m = \dfrac{d}{h}$, and find $c$ from one row.
Otherwise compute the second differences. If they are all equal to $D$, predict a quadratic rule $y = ax^2 + bx + c$ with $a = \dfrac{D}{2h^2}$ ($a = \dfrac{D}{2}$ when $h = 1$). Substitute two rows to obtain two equations in $b$ and $c$, and solve this system.
Check the rule on a row that was not used.
Conclude with care: the rule is a prediction consistent with the table. The table alone does not prove it: a reason taken from the situation is needed, or the knowledge that the rule is linear or quadratic.

Example

The inputs of a table are $x = 0$, $1$, $2$, $3$, $4$ and its outputs are $y = 5$, $8$, $11$, $14$, $17$. Predict a rule giving $y$ in terms of $x$.

Answer

The step is $h = 1$ and the first differences are $3$, $3$, $3$, $3$: all equal to $d = 3$. We predict a linear rule $y = mx + c$ with $m = \dfrac{d}{h} = 3$, and the row $x = 0$ gives $c = 5$: $$y = 3x + 5.$$ Check on $x = 4$: $3 \times 4 + 5 = 17$.

Example

Predict a rule giving $y$ in terms of $x$ for the table below.

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$4$	$9$	$18$	$31$	$48$

Answer

The inputs increase by the same step $h = 1$. The differences, written between the columns they come from:

$x$	$1$		$2$		$3$		$4$		$5$
$y$	$4$		$9$		$18$		$31$		$48$
first		$5$		$9$		$13$		$17$
second			$4$		$4$		$4$

The first differences are not equal; the second differences are all equal to $D = 4$. We predict a quadratic rule $y = ax^2 + bx + c$ with $a = \dfrac{D}{2} = \dfrac{4}{2} = 2$. Substituting the rows $x = 1$ and $x = 2$ into $y = 2x^2 + bx + c$: $$\begin{aligned} 2 + b + c &= 4, && \text{so } b + c = 2 \\ 8 + 2b + c &= 9, && \text{so } 2b + c = 1. \end{aligned}$$ Subtracting the first equation from the second: $b = 1 - 2 = -1$, then $c = 2 - b = 3$. The predicted rule is $$y = 2x^2 - x + 3.$$ Check on the unused row $x = 5$: $2 \times 25 - 5 + 3 = 48$. The rule fits the whole table.

Example

At a meeting, each person shakes hands exactly once with every other person. The table gives the number $H$ of handshakes for $n$ people.

$n$	$2$	$3$	$4$	$5$	$6$
$H$	$1$	$3$	$6$	$10$	$15$

Use the differences to predict a rule giving $H$ in terms of $n$.
Explain why this rule holds for every number $n$ of people.

Answer

The step is $1$. The first differences are $2$, $3$, $4$, $5$; the second differences are all equal to $D = 1$. We predict $H = an^2 + bn + c$ with $a = \dfrac{1}{2}$. The rows $n = 2$ and $n = 3$ give $$\begin{aligned} 2 + 2b + c &= 1, && \text{so } 2b + c = -1 \\ \dfrac{9}{2} + 3b + c &= 3, && \text{so } 3b + c = -\dfrac{3}{2}. \end{aligned}$$ Subtracting: $b = -\dfrac{3}{2} + 1 = -\dfrac{1}{2}$, then $c = -1 - 2b = 0$. The predicted rule is $$H = \dfrac{1}{2}n^2 - \dfrac{1}{2}n = \dfrac{n(n - 1)}{2}.$$ Check on $n = 6$: $\dfrac{6 \times 5}{2} = 15$.
The table alone does not prove the rule. The situation does: each of the $n$ people shakes $n - 1$ hands, which gives $n(n - 1)$ hands shaken; each handshake joins two people, so it has been counted twice. Hence $H = \dfrac{n(n - 1)}{2}$ for every $n$.

A Table Does Not Prove a Rule

Many rules fit the same finite table. The table with inputs $1$ to $5$ and outputs $4$, $9$, $18$, $31$, $48$, fitted above by $y = 2x^2 - x + 3$, is also produced by $$y = 2x^2 - x + 3 + (x - 1)(x - 2)(x - 3)(x - 4)(x - 5),$$ since the product is $0$ for $x = 1, 2, 3, 4, 5$. At $x = 6$, however, this rule gives $69 + 120 = 189$, while $y = 2x^2 - x + 3$ gives $69$. The Method finds the simplest rule of degree at most $2$ consistent with the table: it is a model, to be checked on further values and justified by the situation. The inputs must also be equally spaced: with the inputs $1$, $2$, $4$, the differences of $y = x^2$ are $3$ and $12$, and they cannot be compared directly.

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Skills to practice

Recognising Linear and Quadratic Tables
Finding the Rule from a Table

VII Going further

Skills to practice

Reasoning and Problem-Solving
Solving Problems with Inequalities
Investigating Patterns
Quiz

\(x\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(y\)	\(4\)	\(9\)	\(18\)	\(31\)	\(48\)

\(x\)	\(1\)		\(2\)		\(3\)		\(4\)		\(5\)
\(y\)	\(4\)		\(9\)		\(18\)		\(31\)		\(48\)
first		\(5\)		\(9\)		\(13\)		\(17\)
second			\(4\)		\(4\)		\(4\)

\(n\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(H\)	\(1\)	\(3\)	\(6\)	\(10\)	\(15\)