NCEA Level 1Linear equations and inequationsAS 91947

Solving linear equations and inequations

What this lesson covers

Solving is undoing, with the same operation on both sides, so the balance is kept.
An inequation has a range of solutions, and multiplying or dividing by a negative reverses its sign.

Key idea

What Level 1 adds

You solved one-sided equations and one inequation in Year 10.
Now the unknown sits on both sides, and questions start in words.
And every step needs a reason, not just an answer.

Year 10	Level 1
x on one side	x on both sides, with brackets
One fraction	two fractions with different denominators
An equation given to you	an equation you form from words
One inequation	inequations, double ones, whole-number answers
An answer	an answer with a reason for every step

Worked example

The unknown on both sides

Solve 3(2x − 1) = 4x + 9.
Expand first, then move every x term to one side.
x = 6, and both sides come to 33.

Step	Leaves	Why
Expand the bracket	6x − 3 = 4x + 9	distributive law
Subtract 4x	2x − 3 = 9	same on both sides
Add 3	2x = 12	same on both sides
Divide by 2	x = 6	same on both sides
Check	33 = 33	both sides agree

Worked example

A minus in front of the bracket

Solve 5 − 2(x − 4) = 3x − 7.
The −2 multiplies both terms: minus times minus is plus 8.
Move x to the side where it stays positive. x = 4.

Step	Leaves	Why
Expand −2(x − 4)	5 − 2x + 8 = 3x − 7	−2 × −4 = +8
Collect	13 − 2x = 3x − 7	like terms
Add 2x, add 7	20 = 5x	x term stays positive
Divide by 5	x = 4	same on both sides
Check	5 = 5	5 − 0 and 12 − 7

Key idea

Saying why, not just what

Name the operation: "subtract 4x from both sides".
Both sides get the same step, so the equation stays balanced.
Finish by checking in the original, not your last line.

Worked example

Two fractions: clear them both at once

Solve (x + 1)/3 + (x − 1)/4 = 3.
Times every term by 12, the common denominator.
Keep each numerator in a bracket. x = 5.

Step	Leaves	Note
Common denominator	12	LCM of 3 and 4
Times every term	4(x + 1) + 3(x − 1) = 36	keep the brackets
Expand, collect	7x + 1 = 36
Subtract 1, divide by 7	x = 5
Check	2 + 1 = 3	correct

Key idea

Forming an equation from words

Plan A: $30 a month plus $2 per GB.
Plan B: $45 a month plus 50c per GB.
For how much data do they cost the same?

Stage	What you write
1. Define	let g be the data used in a month, in GB
2. Write each quantity	the cost of each plan, in dollars
3. Link them	"costs the same" means an equals sign
4. Solve and check	then answer the question in words

Worked example

Solving the phone plan

Each plan is a fixed charge plus a rate times g.
Same cost, so set the two expressions equal.
g = 10. Both plans cost $50.

Plan	Cost	Note
Plan A	30 + 2g	dollars
Plan B	45 + 0.5g	dollars
Same cost	30 + 2g = 45 + 0.5g
Subtract 30, 0.5g	1.5g = 15	g = 10
Check	$50 and $50	correct

Worked example

What the answer means

At 10 GB a month, the two plans cost the same.
Below 10 GB, Plan A is cheaper.
Above 10 GB, Plan B is cheaper.

Data	Plan A	Plan B
4 GB	$38	$47
10 GB	$50	$50
20 GB	$70	$55

Worked example

An inequation with a reversal

Solve 3(x + 2) > 5x − 4.
Dividing by −2 reverses the sign: x < 5.
Solve it the other way round and you get x < 5 again.

Step	Leaves	Note
Expand	3x + 6 > 5x − 4
Subtract 5x and 6	−2x > −10
Divide by −2	x < 5	sign reverses
Or move x right	10 > 2x, so x < 5	same answer
Check x = 4	18 > 16	true

Worked example

The answer is a set of numbers

Every number less than 5 works.
5 itself does not: 21 > 21 is false.
On paper: an open circle at 5, shaded to the left.

Worked example

A double inequation

Solve −3 ≤ 2x + 1 < 9.
Do the same step to all three parts.
−2 ≤ x < 4. Dividing by a positive: no reversal.

Step	Leaves	Note
Subtract 1 from all three	−4 ≤ 2x < 8
Divide all three by 2	−2 ≤ x < 4	positive, no reversal
Left end	−2 included	≤
Right end	4 excluded	<

Worked example

Between two ends

On paper: closed circle at −2, as −2 is included.
On paper: open circle at 4, as 4 is not.
Whole numbers only: −2, −1, 0, 1, 2 and 3.

Worked example

How many tickets at most?

$250 for a class trip. Bus $85, tickets $12 each.
t ≤ 13.75, but you cannot buy part of a ticket.
Round down: at most 13. Fourteen goes over.

Step	Leaves	Note
Budget	85 + 12t ≤ 250	t = tickets
Subtract 85	12t ≤ 165
Divide by 12	t ≤ 13.75
Whole tickets	at most 13	round down
Check	13: $241, 14: $253	14 is over

Summary

What to take away

Seven lines, and the first one is the reason for every step.


Solving is undoing	the same step on both sides
Brackets first	a minus outside hits both terms
Two fractions	clear with the common denominator
From words	define the letter, with its units
An inequation	has a range of solutions
× or ÷ by a negative	reverses the sign
In context	round the way the question needs

Your turn

Your turn: four questions

Pause here and write each one out, with reasons.
Question 1: watch the minus in front of the second fraction.
Questions 2 and 4: define your letter first.

	Question
1	Solve (x + 1)/3 − (x − 1)/4 = 1.
2	Hāngī costs $180, plates sell at $15. Plates for $420 profit?
3	Solve 7 − 3x ≥ 19 and show it on a number line.
4	Van $140, $35 per paddler, budget $500. Most paddlers?

Show answer

1. x = 5. Times every term by 12: 4(x + 1) − 3(x − 1) = 12. The minus multiplies both terms, so 4x + 4 − 3x + 3 = 12, x + 7 = 12. Check: 2 − 1 = 1. 2. 40 plates. Let p be the number of plates: 15p − 180 = 420, 15p = 600, p = 40. 3. x ≤ −4. Subtract 7 for −3x ≥ 12, then divide by −3, which reverses the sign. Closed dot at −4, shaded to the left. 4. At most 10 paddlers. 140 + 35n ≤ 500, 35n ≤ 360, n ≤ 10.28… (360 ÷ 35), rounded down because 11 would cost $525.

Worked example

The answers

1. x = 5.
2. 15p − 180 = 420, so 40 plates.
3. x ≤ −4, closed circle at −4, shaded left.
4. n ≤ 10.28… (360 ÷ 35), so at most 10 paddlers.

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