Equations and inequalities (Grade 10) — Mathematics Grades 8–11

Solve linear and factorisable quadratic equations, and linear inequalities.

Gr 10 Term 1 - Equations and inequalitiesTerm 1 - 2 weeks (CAPS)

Practise Equations and inequalities in the app →

What gets asked

Solve a linear equation with brackets or fractions.
Solve a quadratic equation by factorising it first.
Solve two simultaneous equations by substitution or elimination.
Solve a linear inequality and write the answer in interval notation.

You must be able to

Clear a fraction by multiplying every term by the common denominator.
Write a quadratic as bracket times bracket equals zero, then solve each bracket.
Substitute one equation into the other to solve for two unknowns.
Flip the inequality sign when multiplying or dividing by a negative number.

Traps that cost marks

Dividing both sides of an equation like $x^{2} = 5 x$ by $x$ , which throws away one answer. Move everything to one side first: $x^{2} - 5 x = 0$ , then factorise. Never divide by the unknown.
Adding two equations together when the matching terms already have the same sign. Same sign means subtract, not add. Check the signs before you combine two equations.
Keeping the inequality sign the same after dividing both sides by a negative number. Dividing by a negative flips the sign. $- 2 x > 6$ becomes $x < - 3$ .

Worked example

Solve:

x^{2} - 2 x - 15 = 0

Factorise: $(x - 5) (x + 3) = 0$ .
Set each bracket to zero: $x - 5 = 0$ or $x + 3 = 0$ .
$x = 5$ or $x = - 3$ .

Linear equations

Finding the number a letter stands for, even when it is hiding underneath a fraction.

equation: A statement that two expressions are equal, like $2 x + 5 = 11$ .
solution: The number that makes both sides equal. No other number does.
common denominator: A number every denominator goes into exactly. For $4$ and $6$ you can use $12$ .
restriction: A value the letter may not take, because it would make a denominator $0$ .

An equation is a balance. The two sides weigh the same. Whatever you do to the one side, do to the other, or the balance breaks. Solving means peeling operations off the letter until it stands alone. The number left over is the solution, and no other number works.

Brackets come off first. Every term inside gets multiplied, and the sign in front travels with the number. So $- 2 (x - 4)$ is $- 2 x + 8$ , not $- 2 x - 8$ . Miss that sign and the rest of the working is wasted.

Fractions come off next. Multiply every single term by a common denominator, including the terms with no fraction in them. Then $\frac{x}{2} + \frac{x}{3} = 5$ turns into $3 x + 2 x = 30$ , which is easy.

Sometimes the letter sits in the denominator. Write the restriction down before you start: the letter may never make a denominator $0$ . Then multiply both sides by that bracket. At the end, check that your answer is not the barred value.

A word problem needs one extra step at the start. Say in words what the letter stands for. Turn each sentence into maths. When you have the number, read the question again and answer what it actually asked.

Rules to remember

$a (b - c) = ab - a c$
You may multiply or divide both sides by any number except $0$

Examples

Solve:

3 x - 2 (x - 4) = 11

Worked answer

Expand the bracket. The $- 2$ multiplies both terms.
$- 2 (x - 4) = - 2 x + 8$ , so we get $3 x - 2 x + 8 = 11$ .
The left side tidies up to $x + 8 = 11$ .
Subtract $8$ from both sides: $x = 3$ .
Check: $3 (3) - 2 (3 - 4) = 9 + 2 = 11$ .

Answer: $x = 3$

The minus sign in front of the bracket turned the $- 4$ into $+ 8$ .

Solve:

\frac{x + 6}{x - 2} = 3

Worked answer

The denominator may not be $0$ , so the restriction is that $x$ is not $2$ .
Multiply both sides by $x - 2$ : $x + 6 = 3 (x - 2)$ .
Expand the right side: $x + 6 = 3 x - 6$ .
Take $x$ across and add $6$ : $12 = 2 x$ , so $x = 6$ .
Since $6$ is not $2$ , this answer is allowed.

Answer: $x = 6$

The bracket had to multiply every term on the right, not only the first one.

A white loaf costs R2 more than a brown loaf. Three white loaves and two brown loaves cost R86. What does a white loaf cost?

Worked answer

Let $b$ be the price of a brown loaf, in rands.
A white loaf costs $2$ more, so it is $b + 2$ .
Three white and two brown give $3 (b + 2) + 2 b = 86$ .
Expand and collect: $5 b + 6 = 86$ , so $5 b = 80$ and $b = 16$ .
A white loaf was asked for: $16 + 2 = 18$ .

Answer: R18

The letter stood for the brown loaf, so one more step was needed at the end.

Traps

Expanding $3 x - 2 (x - 4)$ as $3 x - 2 x - 8$ . The $- 2$ multiplies both terms inside: $- 2 (x - 4) = - 2 x + 8$ .
Multiplying only the fractions by the common denominator and leaving the whole numbers alone. Every term is multiplied. In $\frac{x}{2} + \frac{x}{3} = 5$ the $5$ becomes $30$ .
Giving $x = 4$ for $\frac{x}{x - 4} = \frac{4}{x - 4}$ , where $4$ is barred. Write the restriction first. Here $x$ may not be $4$ , so there is no solution.
Answering R16 when the question asked what a white loaf costs. The letter stood for the brown loaf. Add the $2$ : $16 + 2 = 18$ .

Learn and practise “Linear equations” in the app →

Changing the subject

Rearranging a formula so that the letter you actually want is the one standing on its own.

formula: A rule written as an equation, such as $P = 2 l + 2 b$ for a perimeter.
subject: The letter standing alone on one side, with nothing else beside it.
common factor: Something that sits in every term. In $a x - c x$ it is $x$ .
square root: $n$ is the number that multiplies by itself to give $n$ .

A formula ties quantities together. The letter you want is not always the one standing alone. Changing the subject means rearranging it until your letter does stand alone. Nothing new is needed. The rules that move things across an equation work here too.

Work backwards, exactly as you would when solving. See what has been done to your letter, then undo it in reverse order. A term that is added gets subtracted from both sides. A number that multiplies gets divided out.

A square needs the whole side. To undo it, take the square root of everything on that side at once. Rooting term by term is wrong, and it is the slip made most often here.

Your letter may sit in more than one term. Move all of those terms onto the same side. Take out the common factor, which leaves a bracket behind. Then divide both sides by that whole bracket.

If the letter is stuck in a denominator, multiply both sides by it first. Keep the restriction in mind: no denominator may be $0$ .

Rules to remember

$a x - c x = x (a - c)$
You may do to a formula anything you may do to an equation

Examples

Make

b

the subject of

P = 2 l + 2 b

Worked answer

The $2 l$ is added on the right, so subtract it from both sides.
That leaves $P - 2 l = 2 b$ .
The $b$ has been multiplied by $2$ , so divide both sides by $2$ .
This gives $b = \frac{P - 2 l}{2}$ .

Answer: $b = \frac{P - 2 l}{2}$

Each operation came off in the reverse order to the one that built the formula.

Make

r

the subject of

A = π r^{2}

Worked answer

Divide both sides by $π$ : $\frac{A}{π} = r^{2}$ .
Now undo the square. Take the square root of that whole side.
So $r = \frac{A}{π}$ .
A radius is never negative, so only the positive root is kept.

Answer: $r = \frac{A}{π}$

The root went over the entire right-hand side, not over $A$ on its own.

Make

x

the subject of

a x - 5 = c x + 7

Worked answer

Both sides carry an $x$ term, so gather them on one side.
Subtract $c x$ and add $5$ : $a x - c x = 12$ .
Take out the common factor: $x (a - c) = 12$ .
Divide by the whole bracket: $x = \frac{12}{a - c}$ .
This needs $a$ and $c$ to differ, or the bracket is $0$ .

Answer: $x = \frac{12}{a - c}$

Dividing by $x$ early would have been useless while a second $x$ was still there.

Traps

Going from $P = 2 l + 2 b$ straight to $P + 2 l = 2 b$ . The $2 l$ is added, so subtract it from both sides: $P - 2 l = 2 b$ .
Turning $y = \frac{k}{x}$ into $x = \frac{y}{k}$ . Multiply by $x$ first: $x y = k$ . Then divide by $y$ to get $x = \frac{k}{y}$ .
Reading $c^{2} = a^{2} + b^{2}$ as $c = a + b$ . Root the whole side at once, so $c = a^{2} + b^{2}$ .
Dividing $a x = c x + 7$ by $x$ while an $x$ is still on the right. Bring every $x$ term to one side first, then take out the common factor.

Learn and practise “Changing the subject” in the app →

Quadratic equations

When the letter is squared there are usually two answers, and brackets are how you find them.

quadratic equation: An equation in which the highest power of the letter is $2$ .
standard form: Every term gathered on one side, with exactly $0$ on the other.
zero-product rule: If two things multiply to give $0$ , at least one of them is $0$ .
root: A value of the letter that makes the equation true.

A quadratic equation carries an $x^{2}$ . That changes the method completely. You cannot peel operations off one at a time, because the letter appears twice over. Instead you build a product that equals $0$ and read the answers off the brackets. There are usually two of them.

Standard form comes first. Shift every term to one side so that the other side is exactly $0$ . Then factorise what is left. A common factor may be all you need.

Now the zero-product rule does the work. Two brackets multiply to give $0$ , so at least one bracket must be $0$ . Set each bracket equal to $0$ in turn. Each one gives a small, easy equation and one root.

The rule only works against $0$ . Two brackets multiplying to give $15$ tell you nothing, because many pairs of numbers give $15$ . So never split brackets while some other number is sitting on the right.

Never divide both sides by the letter, because that throws a root away. In a word problem, test each root against the story. A length, an age or a number of loaves cannot be negative, so one root is often dropped.

Rules to remember

If $A \times B = 0$ then $A = 0$ or $B = 0$
A quadratic equation has at most two roots

Examples

Solve:

x^{2} - 7 x + 12 = 0

Worked answer

Look for two numbers that multiply to $12$ and add to $- 7$ .
They are $- 3$ and $- 4$ , since $- 3 \times - 4 = 12$ .
So the left side factorises: $(x - 3) (x - 4) = 0$ .
Set each bracket to $0$ : $x - 3 = 0$ or $x - 4 = 0$ .
That gives $x = 3$ or $x = 4$ .

Answer: $x = 3$ or $x = 4$

The pair had to multiply to the last number and add to the one in front of $x$ .

Solve:

x (x + 2) = 15

Worked answer

The right side is not $0$ , so the bracket rule cannot be used yet.
Expand the left side: $x^{2} + 2 x = 15$ .
Move the $15$ across: $x^{2} + 2 x - 15 = 0$ .
Factorise: $(x + 5) (x - 3) = 0$ .
So $x = - 5$ or $x = 3$ .

Answer: $x = - 5$ or $x = 3$

Splitting the brackets against $15$ would have been wrong, because $15$ is not $0$ .

A vegetable garden is

3

m longer than it is wide, and covers

40

square metres. How wide is it?

Worked answer

Let the width be $w$ metres, so the length is $w + 3$ .
Area is length times width: $w (w + 3) = 40$ .
Expand and move across: $w^{2} + 3 w - 40 = 0$ .
Factorise: $(w + 8) (w - 5) = 0$ , so $w = - 8$ or $w = 5$ .
A width cannot be negative, so the garden is $5$ m wide.

Answer: $5$ m wide

Area multiplies two changing lengths, so the model was bound to be quadratic.

Traps

Dividing both sides of an equation like $x^{2} = 5 x$ by $x$ , which throws away one answer. Move everything to one side first: $x^{2} - 5 x = 0$ , then factorise. Never divide by the unknown.
From $x (x + 2) = 15$ writing $x = 15$ or $x + 2 = 15$ . The rule needs $0$ on the right. Expand, shift every term across, then factorise.
Moving the $15$ across and writing $x^{2} + 2 x + 15 = 0$ . Subtracting $15$ from both sides leaves $- 15$ , so $x^{2} + 2 x - 15 = 0$ .
Giving a width of $- 8$ m because $- 8$ came out as a root. A width is never negative. Test both roots against the story and drop the impossible one.

Learn and practise “Quadratic equations” in the app →

Simultaneous equations

Two letters at once, and the two facts you need to pin both of them down.

simultaneous equations: Two equations that must both be true, for the same pair of numbers.
substitution: Replacing a letter by something you know is equal to it.
elimination: Wiping out one letter by adding or subtracting two equations.
coefficient: The number in front of a letter. In $3 y$ it is $3$ .

Two unknowns need two facts. One equation on its own has endless pairs that fit it. A second equation cuts those down to a single pair. That pair is the solution, and it must satisfy both simultaneous equations at once.

Substitution is the first method. Make one letter the subject of one equation. Put that expression into the other equation, not back into the one you rearranged. Only one letter is left, and you already know how to solve for it.

Elimination is the second method. Write the two equations under each other, lining up like terms. If one letter has matching terms, add or subtract to wipe it out. Same sign means subtract. Different signs mean add.

Often the terms do not match yet. Multiply a whole equation, every term of it, by a number that makes the coefficient you want match the other one. Multiplying only one side breaks the balance.

Whichever method you pick, finish the job. Put the number you found back to get the second one. Then check both equations. In a word problem, two different quantities need two different letters.

Rules to remember

Same sign: subtract. Different signs: add
You may multiply a whole equation by any number except $0$

Examples

Solve together:

y = x + 2

and

3 x + y = 14

Worked answer

The first equation already has $y$ standing alone.
Put $x + 2$ where $y$ is in the second: $3 x + (x + 2) = 14$ .
That tidies to $4 x + 2 = 14$ , so $4 x = 12$ .
Dividing by $4$ gives $x = 3$ .
Now find $y$ : $y = 3 + 2 = 5$ .

Answer: $x = 3$ and $y = 5$

The expression went into the other equation, which left one letter to solve.

Solve together:

3 x + 2 y = 23

and

x + 2 y = 13

Worked answer

Both equations carry $+ 2 y$ . Same sign, so subtract.
The $y$ terms cancel: $3 x - x = 2 x$ and $23 - 13 = 10$ .
So $2 x = 10$ , which gives $x = 5$ .
Put $x = 5$ into the second equation: $5 + 2 y = 13$ .
Then $2 y = 8$ , so $y = 4$ .

Answer: $x = 5$ and $y = 4$

Subtracting was right because the two $y$ terms already had the same sign.

At a spaza shop, two loaves and three litres of milk cost R102. One loaf and two litres cost R62. Find both prices.

Worked answer

Let $b$ be the price of a loaf and $m$ the price of a litre of milk.
The two sentences give $2 b + 3 m = 102$ and $b + 2 m = 62$ .
Make $b$ the subject of the second: $b = 62 - 2 m$ .
Substitute into the first: $2 (62 - 2 m) + 3 m = 102$ , so $124 - m = 102$ .
Then $m = 22$ , and $b = 62 - 44 = 18$ .

Answer: a loaf costs R18 and a litre of milk costs R22

Two different goods needed two different letters, and two sentences gave two equations.

Traps

Adding two equations together when the matching terms already have the same sign. Same sign means subtract, not add. Check the signs before you combine two equations.
Doubling only the left side of an equation to make the coefficients match. Multiply every term, on both sides. Doubling $x + 2 y = 13$ gives $2 x + 4 y = 26$ .
Stopping at $x = 5$ and never working out the second letter. A pair of equations has a pair of answers. Put $x = 5$ back in to find $y$ .
Using $x$ for both the bread price and the milk price. Two different quantities need two different letters, or the second equation says nothing new.

Learn and practise “Simultaneous equations” in the app →

Linear inequalities

When the answer is not one number but a whole stretch of them, and three ways to write it down.

inequality: A statement that one side is bigger or smaller than the other.
interval notation: A short way to write a stretch of numbers, using its two end values.
compound inequality: Two inequalities on one line, such as $- 3 < 2 x + 1 \leq 9$ .
infinity: Not a number. It says the values run on without ever stopping.

An equation usually has one solution. An inequality has a whole stretch of them. Solving one means describing that stretch, not landing on a single number. There are three ways to write the answer down: in symbols, on a number line, and in interval notation.

Solve it exactly as you solve an equation, with one extra rule. Multiply or divide both sides by a negative number and the sign turns around. Adding and subtracting never flip it.

On a number line, mark the end value. Use a filled dot when that value counts, which is $\leq$ or $\geq$ . Use an open dot when it does not, which is $<$ or $>$ . Then shade towards the values that work.

Interval notation puts the two ends in brackets, with a semicolon between them. A square bracket takes the end value in. A round bracket leaves it out. Against infinity the bracket is always round, because that end is never reached.

A compound inequality holds the letter in the middle. Do the same thing to all three parts, every single time. Working on the middle alone breaks the statement apart.

Rules to remember

You may add or subtract the same number on every part
Multiplying or dividing by a negative number turns the sign around

Examples

Solve, and write the answer in interval notation:

3 x - 5 \geq 7

Worked answer

Add $5$ to both sides: $3 x \geq 12$ .
Divide both sides by $3$ . It is positive, so the sign stays.
That gives $x \geq 4$ .
Number line: a filled dot at $4$ , shaded to the right.
Interval notation: $[4; \infty)$ .

Answer: $x \geq 4$ , or $[4; \infty)$

Dividing by a positive number left the sign exactly as it was.

Solve:

4 - 3 x \leq 19

Worked answer

Subtract $4$ from both sides: $- 3 x \leq 15$ .
Now divide both sides by $- 3$ , which is negative.
So the sign turns around: $x \geq - 5$ .
Number line: a filled dot at $- 5$ , shaded to the right.

Answer: $x \geq - 5$

The flip came only at the division, not when the $4$ moved across.

Solve, and write the answer in interval notation:

- 3 < 2 x + 1 \leq 9

Worked answer

There are three parts here. Subtract $1$ from all three.
That leaves $- 4 < 2 x \leq 8$ .
Divide all three parts by $2$ , which is positive.
So $- 2 < x \leq 4$ .
Interval notation: $(- 2; 4]$ .

Answer: $- 2 < x \leq 4$ , or $(- 2; 4]$

Every part changed together, so the letter stayed trapped between the same two ends.

Traps

Keeping the inequality sign the same after dividing both sides by a negative number. Dividing by a negative flips the sign. $- 2 x > 6$ becomes $x < - 3$ .
Drawing a filled dot at $- 3$ for $x < - 3$ . A filled dot means the value counts. For $<$ the dot is left open.
Writing the answer as $[4; \infty]$ . You never arrive at infinity, so that bracket stays round: $[4; \infty)$ .
Taking $1$ off only the middle part of $- 3 < 2 x + 1 \leq 9$ . All three parts change together, which gives $- 4 < 2 x \leq 8$ .

Learn and practise “Linear inequalities” in the app →

Grade 10 Equations and inequalities

What gets asked

You must be able to

Traps that cost marks

Worked example

Linear equations

Rules to remember

Examples

Traps

Changing the subject

Rules to remember

Examples

Traps

Quadratic equations

Rules to remember

Examples

Traps

Simultaneous equations

Rules to remember

Examples

Traps

Linear inequalities

Rules to remember

Examples

Traps

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