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

Solving linear equations, and setting one up from a word problem.

Gr 8 Terms 1, 2 and 4 — Algebraic equationsTerm 1, Term 2 and Term 4

Practise Algebraic equations in the app →

What gets asked

Solve a one-step or two-step linear equation.
Set up an equation to describe a word problem.
Solve a word problem by forming and solving an equation.
Use a rule to complete a table of input and output pairs.

You must be able to

Do the same operation to both sides of an equation to keep it balanced.
Remove the constant term before dividing by the coefficient.
Choose one variable to represent the unknown in a word problem.
Answer the actual question asked, not just solve for the variable.

Traps that cost marks

Dividing by the coefficient before the constant term has been removed. Remove the constant term first (add or subtract), and only then divide by the coefficient.
Stopping at $- x = 8$ instead of dividing both sides by $- 1$ . A solution cannot have a negative sign in front of the letter. Divide both sides by $- 1$ : $x = - 8$ .
Solving for the chosen variable but answering with the wrong person's or item's amount. Check what the question actually asks for before you write your final answer.
Writing an expression with no equal sign where a full equation is needed. An equation always needs an equal sign linking two sides.

Worked example

Solve:

3 x + 4 = 19

Subtract 4 from both sides: $3 x = 15$ .
Divide both sides by 3: $x = 5$ .
Check: $3 (5) + 4 = 19$ , which matches.

Solve equations

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

equation: Two sides joined by an equals sign, such as $3 x + 4 = 19$ .
solution: The number that makes both sides equal. No other number does.
inverse: The operation that undoes another. Subtracting undoes adding, and dividing undoes multiplying.
inspection: Solving by asking what number would fit, then checking it, instead of working step by step.
coefficient: The number in front of the letter. In $5 x$ the coefficient is $5$ .
constant term: A term that is only a number, with no letter on it. In $3 x + 4$ the constant term is $4$ .

An equation is a scale that balances. The two sides weigh the same. Whatever you do to one side you must do to the other, or it tips. That one rule is the whole method.

Some you can just see. In $n + 7 = 12$ the answer is $5$ , because $5 + 7 = 12$ . That is solving by inspection. Put your answer back into the equation and check it.

Otherwise undo each operation with its inverse. In $7 m = 42$ the $m$ was multiplied by $7$ , so divide both sides by $7$ , giving $m = 6$ . If you land on $- x = 8$ , divide both sides by $- 1$ to get $x = - 8$ .

With two steps, move the constant term first, then divide by the coefficient. In $5 x - 3 = 12$ , add $3$ to both sides to get $5 x = 15$ . Then divide both sides by $5$ , so $x = 3$ .

An exponential equation has the letter up in the exponent. In $2^{x} = 32$ you ask how many $2$ s multiply together to make $32$ . Since $2^{5} = 32$ , the solution is $x = 5$ .

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 $0$
Undo the last thing that was done to the letter, first

Examples

Solve by inspection:

n + 7 = 12

Worked answer

Ask what number added to $7$ gives $12$ .
Try $5$ . Then $5 + 7 = 12$ , which fits.
So $n = 5$ .

Answer: $n = 5$

The answer wanted is the value of $n$ , not the value of the whole side.

Solve:

4 t - 5 = 19

Worked answer

Move the constant term first: add $5$ to both sides.
That gives $4 t = 24$ .
Now divide both sides by the coefficient $4$ .
So $t = 6$ .
Check: $4 \times 6 - 5 = 19$ .

Answer: $t = 6$

Clearing the $- 5$ first left one plain multiplication to undo.

Solve:

3^{x} = 81

Worked answer

$3^{x}$ means that many $3$ s multiplied together, not $3 \times x$ .
Count up: $3^{2} = 9$ , then $3^{3} = 27$ , then $3^{4} = 81$ .
The bases now match, so the exponents match too.
So $x = 4$ .

Answer: $x = 4$

What you are hunting for is the exponent, not the answer to a multiplication.

Traps

Dividing by the coefficient before the constant term has been removed. Remove the constant term first (add or subtract), and only then divide by the coefficient.
Stopping at $- x = 8$ instead of dividing both sides by $- 1$ . A solution cannot have a negative sign in front of the letter. Divide both sides by $- 1$ : $x = - 8$ .
Answering $12$ for $n + 7 = 12$ , which is what the whole side is worth. The question asks for $n$ on its own. Here $n = 5$ .
Reading $2^{x} = 32$ as $2 x = 32$ and answering $16$ . The $x$ is an exponent. Since $2^{5} = 32$ , the solution is $x = 5$ .

Learn and practise “Solve equations” in the app →

Set up and solve problems

Turning a story about money, or about a garden, into an equation you can solve.

variable: A letter standing for a number you do not know yet, or one that can change.
formula: A rule written as an equation, such as $A = l \times b$ for the area of a rectangle.
perimeter: The total distance right round the outside of a shape.
area: How much flat space a shape covers, counted in squares such as square metres.

An equation can describe something real. A school feeding scheme buys maize meal, and the cost in rands is $c = 95 n + 60$ for $n$ bags. The $95$ is the price of one bag. The $60$ is the delivery, which stays the same however many bags are bought.

To build one yourself, choose a letter for what you do not know. Write down exactly what the words say. Then check that there is an equals sign, because an expression on its own is not an equation.

When two amounts are linked, use one letter for both. If Ayanda made $x$ rands and Lerato made R30 more, then Lerato made $x + 30$ . Two different letters would leave you with two unknowns and no way through.

Solve, then read the question again. The working may hand you Ayanda's amount when Lerato's was asked for. Your last line must answer the question that was actually put.

Shapes give equations too. Perimeter adds up the sides all the way round. Area of a rectangle multiplies length by breadth. Use the formula, not whichever numbers happen to be printed.

Rules to remember

$A = l \times b$ for a rectangle, with length $l$ and breadth $b$
An equation always needs an equals sign

Examples

The cost of maize meal is

c = 95 n + 60

rands for

n

bags. What do

95

and

60

stand for? Find

c

when

n = 20

Worked answer

The $95$ is the price of one bag, in rands.
The $60$ is the delivery charge, which never changes.
Put $20$ in place of $n$ : $95 \times 20 + 60 = 1960$ .

Answer: R $95$ a bag, R $60$ delivery; $c = 1960$

The number attached to $n$ grows with the order. The lone number does not.

Ayanda and Lerato sell vetkoek and make R150 together. Lerato makes R30 more than Ayanda. How much does Lerato make?

Worked answer

Let Ayanda make $x$ rands. Then Lerato makes $x + 30$ .
Together they make R150, so $x + x + 30 = 150$ .
That is $2 x + 30 = 150$ , so $2 x = 120$ .
Divide both sides by $2$ , giving $x = 60$ .
Ayanda made R60, so Lerato made R90.

Answer: Lerato makes R $90$

The solving gave Ayanda's amount, but the question asked about Lerato.

A rectangular vegetable garden is

7

m long and covers

28

square metres. How wide is it?

Worked answer

Area of a rectangle is length times breadth.
Let the breadth be $b$ metres, so $7 b = 28$ .
Divide both sides by $7$ , giving $b = 4$ .
Check: $7 \times 4 = 28$ .

Answer: $4$ m wide

The area formula built the equation. Adding the two numbers would have built nothing.

Traps

Writing an expression with no equal sign where a full equation is needed. An equation always needs an equal sign linking two sides.
Solving for the chosen variable but answering with the wrong person's or item's amount. Check what the question actually asks for before you write your final answer.
Reading the $60$ in $c = 95 n + 60$ as the answer to the whole problem. The $60$ is one fixed part of the cost. The total still has to be worked out.
Adding $7$ and $28$ instead of using the area to find the missing side. Area is length times breadth, so $7 b = 28$ and $b = 4$ .

Learn and practise “Set up and solve problems” in the app →

Equations and tables

Feeding numbers through a rule to build a table, and working backwards when the answer is given.

input: The number you put into a rule. It is usually the $x$ -value.
output: The number that comes out of the rule. It is usually the $y$ -value.
table of values: A table with the inputs written in one row and their outputs in the row underneath.
ordered pair: Two numbers written together in a fixed order, input first: $(2; 5)$ .

A rule like $y = 3 x - 1$ turns every input into an output. Take an input of $2$ and work it out: $3 \times 2 - 1 = 5$ . So an input of $2$ gives an output of $5$ .

Do that for several inputs and set them out in a table of values. The inputs go in the top row and their outputs go underneath. Each column is then an ordered pair, and the input is written first.

A negative input needs brackets, or its sign gets lost. In $y = - 3 x + 2$ with an input of $- 4$ : $- 3 \times (- 4) + 2 = 14$ . Two minus signs multiplied together give a plus.

Sometimes the output is given and the input is missing. Write the rule as an equation and undo it with inverse operations. Running the rule forwards again will not find the input.

Rules to remember

An ordered pair is written input first: $(2; 5)$
Undo a rule with inverse operations, in reverse order

Examples

Build a table of values for

y = 4 x - 3

at inputs

0

1

and

5

Worked answer

With an input of $0$ : $4 \times 0 - 3 = - 3$ .
With an input of $1$ : $4 \times 1 - 3 = 1$ .
With an input of $5$ : $4 \times 5 - 3 = 17$ .

Answer: outputs $- 3$ , $1$ and $17$

One rule was used three times. Only the input changed each time.

For

y = - 2 x + 5

, find the output when the input is

- 3

, then write the ordered pair

Worked answer

Put the $- 3$ inside brackets: $- 2 \times (- 3) + 5$ .
A minus times a minus gives a plus, so $- 2 \times (- 3) = 6$ .
Then $6 + 5 = 11$ .
The input goes first, so the ordered pair is $(- 3; 11)$ .

Answer: $(- 3; 11)$

The brackets kept the minus sign attached to the $3$ .

y = 5 x + 8

the output is

43

. What was the input?

Worked answer

Write it as an equation: $5 x + 8 = 43$ .
Subtract $8$ from both sides: $5 x = 35$ .
Divide both sides by $5$ , giving $x = 7$ .
Check: $5 \times 7 + 8 = 43$ .

Answer: $x = 7$

The rule was undone backwards: the adding went first, then the multiplying.

Traps

Losing the minus when a negative input goes into a rule such as $y = - 3 x + 2$ . Use brackets. With an input of $- 4$ : $- 3 \times (- 4) + 2 = 14$ .
Writing the ordered pair the wrong way round, with the output first. The input always comes first. Input $- 3$ and output $11$ is written $(- 3; 11)$ .
Running the rule forwards again when the output is what you were given. Use the inverse operations instead. From $5 x + 8 = 43$ , subtract first and then divide.

Learn and practise “Equations and tables” in the app →

Grade 8 Algebraic equations

What gets asked

You must be able to

Traps that cost marks

Worked example

Solve equations

Rules to remember

Examples

Traps

Set up and solve problems

Rules to remember

Examples

Traps

Equations and tables

Rules to remember

Examples

Traps

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