Algebraic equations (Grade 9) — Mathematics Grades 8–11

Solving linear equations, setting up an equation from a word problem, and solving by factorising.

Gr 9 Terms 1 and 3 — Algebraic equationsTerm 1 and Term 3

Practise Algebraic equations in the app →

What gets asked

Solve a linear equation, expanding brackets first if needed.
Solve an equation that contains an algebraic fraction.
Set up an equation from a word problem and solve it.
Solve an equation of the form (factor) times (factor) equals 0.

You must be able to

Use inverse operations on both sides of an equation to isolate x.
Multiply every term by the LCM of the denominators to clear fractions.
Choose a variable and write a word problem as an equation before solving.
Set each factor equal to zero and solve each small equation.

Traps that cost marks

Applying an inverse operation to one side of the equation only. Whatever you do to one side of an equation, you must also do to the other side.
Multiplying only the fraction terms by the LCM, and leaving the rest unchanged. Multiply every single term in the equation by the LCM, not just the fraction terms.
Giving only one of the two solutions of a factorised equation. If $(x - 2) (x + 5) = 0$ , either bracket could be zero, so there are two solutions.

Worked example

Solve for x:

3 (x - 2) = x + 8

Expand the bracket: $3 x - 6 = x + 8$ .
Collect x-terms on one side: $2 x = 14$ .
Divide by 2: $x = 7$ .

Solve linear equations

Working backwards to find the number a letter is standing for.

equation: Two expressions joined by an equals sign, such as $3 x + 4 = 19$ .
solution: The number that makes the two sides equal. Every other number does not.
inverse: The operation that undoes another. Subtracting undoes adding; dividing undoes multiplying.
LCM: The smallest number every denominator divides into. For $2$ and $3$ it is $6$ .

An equation is a balance. Whatever you do to one side you must do to the other, or the two sides stop being equal. This is the one rule the whole method rests on.

Undo the operations in reverse order. In $3 x + 4 = 19$ the $x$ was multiplied by $3$ , then $4$ was added. So subtract $4$ first, then divide by $3$ . That gives $3 x = 15$ , then $x = 5$ .

Expand any brackets before you start undoing. Multiply every term inside, not just the first: $3 (x - 2)$ is $3 x - 6$ .

For fractions, multiply every term by the LCM of the denominators. Every term, including the ones with no fraction in them, or the balance is broken.

Always check. Put your solution back into the original equation and see whether the two sides really are equal.

Rules to remember

You may add or subtract the same number on both sides
You may multiply or divide both sides by any number except $0$

Examples

Solve:

4 s = 32

Worked answer

$4 s$ means $4 \times s$ , not $4 + s$ .
The inverse of multiplying by $4$ is dividing by $4$ .
Divide both sides by $4$ : $s = 8$ .
Check: $4 \times 8 = 32$ .

Answer: $s = 8$

Reading $4 s$ as $4 + s$ would give $28$ , and $4 \times 28$ is not $32$ .

Solve:

3 (x - 2) = 18

Worked answer

Expand first: $3 (x - 2) = 3 x - 6$ .
So the equation is $3 x - 6 = 18$ .
Add $6$ to both sides: $3 x = 24$ .
Divide both sides by $3$ : $x = 8$ .
Check: $3 (8 - 2) = 3 \times 6 = 18$ .

Answer: $x = 8$

Expanding $3 (x - 2)$ as $3 x - 2$ would give $x$ a value that fails the check.

Solve:

\frac{x}{2} + \frac{x}{3} = 10

Worked answer

The denominators are $2$ and $3$ , so the LCM is $6$ .
Multiply every term by $6$ , including the $10$ .
$6 \times \frac{x}{2} = 3 x$ and $6 \times \frac{x}{3} = 2 x$ and $6 \times 10 = 60$ .
So $3 x + 2 x = 60$ , which is $5 x = 60$ .
Divide both sides by $5$ : $x = 12$ .

Answer: $x = 12$

Multiplying only the fraction terms by 6 would leave the two sides unbalanced.

Traps

Reading $4 s = 32$ as $4 + s = 32$ and answering $28$ . A number written against a letter means multiply. $4 s$ is $4 \times s$ , so $s = 8$ .
Expanding $3 (x - 2)$ as $3 x - 2$ . Every term inside gets multiplied: $3 (x - 2) = 3 x - 6$ .
Multiplying only the fractions by the LCM and leaving the whole numbers alone. Every term is multiplied. In $\frac{x}{2} + \frac{x}{3} = 10$ the $10$ becomes $60$ too.

Learn and practise “Solve linear equations” in the app →

Exponent and factor equations

Three kinds of equation that do not come apart by simply undoing the operations.

base: The number that carries the exponent. In $2^{x}$ the base is $2$ .
factorise: To rewrite an expression as things multiplied together, inside brackets.
common factor: Something every term can be divided by. In $x^{2} - 3 x$ both terms hold an $x$ .

Most equations come apart with inverse operations. Three kinds do not. One hides the letter up in the exponent. One is already a product. One has to be factorised before anything else can happen.

For an exponent equation, write both sides with the same base. Once the bases match, the exponents must match too. Since $32 = 2^{5}$ , the equation $2^{x} = 32$ gives $x = 5$ .

If two things multiply to give $0$ , at least one of them has to be $0$ . So when one bracket times another equals zero, set each bracket to $0$ on its own. That gives two solutions, and both of them count.

Do not expand a product that already equals $0$ . Those brackets are the work half done. Multiplying them out puts the question straight back where it started.

When one side is not a product yet, factorise it. In $x^{2} - 3 x = 0$ take out the common factor $x$ . Never divide both sides by $x$ , because that quietly throws away the solution $x = 0$ .

Rules to remember

If $ab = 0$ then $a = 0$ or $b = 0$
With the same base, equal powers mean equal exponents

Examples

Solve:

2^{x} = 32

Worked answer

Write $32$ as a power of $2$ : $32 = 2^{5}$ .
Now both sides carry the base $2$ .
Equal powers of the same base need equal exponents.
So $x = 5$ .
Check: $2^{5} = 32$ .

Answer: $x = 5$

Nothing could be compared until both sides were written with one base.

Solve:

(x - 2) (x + 5) = 0

Worked answer

A product is $0$ only when one of the brackets is $0$ .
Set the first bracket to $0$ : $x - 2 = 0$ , so $x = 2$ .
Set the second bracket to $0$ : $x + 5 = 0$ , so $x = - 5$ .
Both values are solutions, so write down both.

Answer: $x = 2$ or $x = - 5$

Each bracket got its own turn, so neither solution went missing.

Solve:

x^{2} - 3 x = 0

Worked answer

Do not divide by $x$ , and do not expand anything.
Take out the common factor: $x (x - 3) = 0$ .
Now one of the two factors has to be $0$ .
So $x = 0$ or $x - 3 = 0$ .
That gives $x = 0$ or $x = 3$ .

Answer: $x = 0$ or $x = 3$

Factorising kept both solutions, and one of them was zero.

Traps

Comparing the exponents while the two bases are still different. Write both sides to the same base first. $32 = 2^{5}$ , so $2^{x} = 32$ gives $x = 5$ .
Not seeing that $\frac{1}{4}$ is also a power of $2$ . $\frac{1}{4} = 2^{- 2}$ , so $2^{x} = \frac{1}{4}$ gives $x = - 2$ .
Giving only one of the two solutions of a factorised equation. If $(x - 2) (x + 5) = 0$ , either bracket could be zero, so there are two solutions.
Dividing both sides of $x^{2} - 3 x = 0$ by $x$ , and answering $3$ alone. Factorise instead: $x (x - 3) = 0$ , so $x = 0$ or $x = 3$ .

Learn and practise “Exponent and factor equations” in the app →

Substitute and interpret

Putting numbers into an equation, filling in a table from it, and writing one for a real situation.

variable: A letter standing for a number that is allowed to change.
substitute: To put a number in place of a letter, and then work the answer out.
dependent variable: The one you work out from the other. In $y = 4 x + 1$ you choose $x$ , and $y$ follows.
ordered pair: One $x$ -value and its $y$ -value, always written with the $x$ -value first.

To substitute is to put a number where a letter was. Write brackets in the letter's place, drop the number inside, then work it out. Those brackets are what keep a minus sign attached to the number.

In an equation like $y = 4 x + 1$ , the letter $y$ is the dependent variable. You choose a value for $x$ and $y$ follows from it. Each $x$ you pick gives exactly one $y$ .

A table is only several substitutions in a row. Each row of the table is one ordered pair. Write the $x$ -value first and the $y$ -value after it, every time.

Sometimes the table hands you $y$ and leaves the $x$ blank. Then put that value in for $y$ and solve the equation for $x$ , using inverse operations as usual.

To write your own equation, first say in words what your variable stands for. Then test it on numbers you already know. Afterwards read it back and say what each number in it does.

Examples

Find

y

when

x = - 2

, given

y = 2 x^{2} - 3

Worked answer

Write brackets where $x$ was: $2 (- 2)^{2} - 3$ .
Do the power first: $(- 2)^{2} = 4$ .
Then multiply: $2 \times 4 = 8$ .
Then subtract: $8 - 3 = 5$ .
So $y = 5$ .

Answer: $y = 5$

The brackets held the minus with the $2$ , so the square came out positive.

A table for

y = 4 x + 1

has

y = 13

and the

x

-value missing. Find it.

Worked answer

Put $13$ in for $y$ : $4 x + 1 = 13$ .
Subtract $1$ from both sides: $4 x = 12$ .
Divide both sides by $4$ : $x = 3$ .
So the ordered pair for that row starts with $3$ .
Check: $4 \times 3 + 1 = 13$ .

Answer: $x = 3$

The value given was a $y$ -value, so it went in on the $y$ side and left an equation.

A metered taxi charges R14 when you get in, then R8 for every kilometre. Write an equation for the cost.

Worked answer

Let $k$ stand for the number of kilometres travelled.
Let $c$ stand for the cost in rands.
Every kilometre costs R8, so the distance part is $8 k$ .
The R14 is paid once, so add it on: $c = 8 k + 14$ .
Read it back: $8$ is charged per kilometre, $14$ is charged once.

Answer: $c = 8 k + 14$

What is paid once gets added, and what is paid each time gets multiplied.

Traps

Leaving the brackets out, so $2 x^{2}$ at $x = - 2$ comes to $- 8$ . Write the brackets first: $2 (- 2)^{2} = 8$ .
Stopping when the table gives the $y$ -value and leaves $x$ blank. Put the given value in for $y$ , then solve the equation for $x$ .
Writing "Thandi is $6$ years older than Sipho" as $T = S - 6$ . Older means more, so it must add: $T = S + 6$ . Test it with two real ages.
Reading $2 x$ as the number of sweets rather than what they cost. Say what the letter counts. If $x$ is sweets at R2 each, then $2 x$ is the cost.

Learn and practise “Substitute and interpret” in the app →

Grade 9 Algebraic equations

What gets asked

You must be able to

Traps that cost marks

Worked example

Solve linear equations

Rules to remember

Examples

Traps

Exponent and factor equations

Rules to remember

Examples

Traps

Substitute and interpret

Examples

Traps

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