Linear equations, inequalities & graphs

Start here: two ideas, one flow

This topic puts three connected ideas on one page:

Solve an equation (find the value of $x$ that makes it true).
Solve an inequality (same moves, but with one extra rule about flipping).
Graph a linear relation (plot the set of $(x, y)$ pairs that satisfy $y = mx + c$ ).

Equations give one answer; inequalities give a range; graphs show the whole relationship at once. The algebra is the same in all three — only the output differs.

What you will learn

solve linear equations of the form $ax + b = c$ and $ax + b = cx + d$ ,
verify a solution by substitution,
solve one-variable linear inequalities and sketch the solution on a number line,
construct a table of values and plot a linear relation $y = mx + c$ ,
read and interpret the $y$ -intercept $c$ (starting value) and gradient $m$ (rate of change),
use linear relations to model practical problems (taxi fares, pay rates, tank drainage).

1. Solving linear equations

Apply inverse operations to both sides to isolate the variable.

Worked example E Very easy: one operation to undo

Solve $\;x + 7 = 12$ .

The left-hand side says ” $x$ plus $7$ ”. To get $x$ alone, undo the “plus $7$ ” by subtracting $7$ from both sides:

x + 7 - 7 = 12 - 7, \qquad x = 5.

Check: $5 + 7 = 12$ . ✓

Worked example 1 Pronumeral on one side

Solve $\;4x - 5 = 23$ .

Add $5$ : $4x = 28$ .
Divide by $4$ : $x = 7$ .

Check: $4(7) - 5 = 28 - 5 = 23$ . ✓

Worked example 2 Pronumeral on both sides

Solve $\;5x + 4 = 2x + 19$ .

Subtract $2x$ from both sides: $3x + 4 = 19$ .
Subtract $4$ : $3x = 15$ .
Divide by $3$ : $x = 5$ .

Check: $5(5) + 4 = 29$ and $2(5) + 19 = 29$ . ✓

Worked example 3 With brackets and fractions

Solve $\;\dfrac{2x - 1}{3} = 5$ .

Multiply both sides by $3$ : $2x - 1 = 15$ . Then $2x = 16$ , $x = 8$ .

2. Linear inequalities

An inequality uses $<$ , $\leq$ , $>$ or $\geq$ instead of $=$ . The same algebra works, with one twist.

Worked example E2 Very easy: inequality in one step

Solve $\;x + 3 > 10$ .

Subtract $3$ from both sides: $x > 7$ . No flip needed because we did not multiply or divide by a negative.

Worked example 4 Solve an inequality

Solve $\;3x - 4 \leq 11$ .

Add $4$ : $3x \leq 15$ .
Divide by $3$ (positive, no flip): $x \leq 5$ .

On a number line, draw a filled circle at $5$ and shade to the left.

3. Linear relations on the Cartesian plane

A relation of the form $y = mx + c$ has a straight-line graph.

Reading a linear graph

Standard form

y = mx + c.

$c$ is the $y$ -intercept: where the line crosses the $y$ -axis (the value of $y$ when $x = 0$ ).
$m$ is the gradient (slope): the change in $y$ per unit change in $x$ .

Table of values

Pick a few $x$ -values, compute $y$ , plot the points, join them with a straight line.

Worked example E3 Very easy: reading m and c straight from the equation

What are the gradient and $y$ -intercept of $y = 3x + 5$ ?

Compare with $y = mx + c$ : the gradient is $m = 3$ and the $y$ -intercept is $c = 5$ (the line crosses the $y$ -axis at $(0, 5)$ ).

Worked example 5 Plotting y = 2x - 1

Table of values:

$x$	$-1$	$0$	$1$	$2$	$3$
$y$	$-3$	$-1$	$1$	$3$	$5$

$y$ -intercept: $(0, -1)$ - the line crosses the $y$ -axis at $-1$ .
Gradient: $2$ - for every $1$ unit right, $y$ rises by $2$ .

Graph of y = 2x - 1. The line crosses the y-axis at -1 (the intercept c); the highlighted triangle shows the gradient m = rise/run = 2/1 = 2.

Three lines with different gradients and intercepts. Steeper gradient = steeper line. The y-intercept is where each line crosses the vertical axis.

Notice: a positive gradient ( $m > 0$ ) tilts upward left-to-right; a negative gradient ( $m < 0$ ) tilts downward. A larger $|m|$ makes the line steeper.

4. Modelling with a linear relation

Many everyday situations are linear: a start-up fee plus a rate per unit.

Worked example 6 Taxi fare

A taxi charges a flag-fall of $4 plus $2.50 per kilometre.

Fare formula: $F = 4 + 2.5k$ where $k$ is the distance in km.
$y$ -intercept (flag-fall): $4.
Gradient: $2.50 per km.
A $12$ km trip costs $4 + 2.5 \times 12 = 4 + 30 = 34$ dollars.

From a graph of $F$ against $k$ , a $20 fare corresponds to where the line hits $F = 20$ , i.e. $4 + 2.5k = 20$ , so $k = 6.4$ km.

Practice: Year 8 core

Fluency

Solve linear equations

Solve $x + 9 = 15$ .
Solve $3x = 18$ .
Solve $2x + 5 = 17$ .
Solve $3y - 8 = 13$ .
Solve $\dfrac{x}{4} + 3 = 9$ .
Solve $4(a - 2) = 20$ .
Solve $5x - 3 = 2x + 12$ .
Solve $3y + 4 = 7y - 16$ .
Solve $\dfrac{3x + 1}{2} = 8$ .
Solve $\dfrac{x}{3} - \dfrac{x}{4} = 2$ .

Fluency

Solve inequalities

Solve $x + 4 > 10$ .
Solve $2y - 3 \leq 5$ .
Solve $3x + 1 \geq 13$ .
Solve $-4x < 20$ .
Solve $5 - 2x > 1$ .
Solve $\dfrac{y}{2} + 1 \geq 3$ .

Fluency

Tables and graphs

Complete the table for $y = 3x - 2$ :

$x$ $-1$ $0$ $1$ $2$ $3$
$y$ ? ? ? ? ?
Find the $y$ -intercept of $y = -2x + 7$ .
Find the gradient of $y = 4x - 1$ .
Find the $x$ -intercept of $y = 2x - 6$ (where $y = 0$ ).
A line passes through $(0, 3)$ and $(2, 9)$ . Find the gradient.
Does the point $(3, 5)$ lie on $y = 2x - 1$ ? Show by substitution.

$x$	$-1$	$0$	$1$	$2$	$3$
$y$	?	?	?	?	?

Reasoning

Explain and spot the mistake

Kai solves $-3x = 12$ as $x = 4$ . Is Kai correct? Explain.
Mira writes ” $-2x < 6$ so $x < -3$ ”. What has Mira forgotten?
Explain why the graphs of $y = 2x$ and $y = 2x + 3$ are parallel.
Describe what happens to the graph of $y = mx + c$ as you increase $c$ .

Problem solving

Modelling

A gym charges a $50 joining fee and $15 per week. Write a formula $C = 50 + 15w$ for cost after $w$ weeks. After how many weeks does the total pass $200?
A water tank starts with $500$ L and leaks at $8$ L per hour. Write a formula for the volume $V$ after $t$ hours. When is the tank empty?
A taxi has flag-fall $3.80 and charges $2 per km. A trip costs $19.80. How long was it (in km)?
A mobile plan charges $30 per month plus $0.05 per text. Lucy’s monthly bill was $33.50. How many texts did she send?
Two phone plans: Plan A charges $20 + $0.15/min; Plan B charges $35 + $0.05/min. Write formulas. For how many minutes are the two plans equal in cost?

Challenge

Reasoning

Harder reasoning

Solve simultaneously (by substitution): $y = 2x - 1$ and $y = x + 4$ .
Solve $\dfrac{x + 1}{2} - \dfrac{x - 1}{3} = 2$ .
A line has gradient $3$ and passes through $(2, 7)$ . Find the equation of the line.
The sum of three consecutive even numbers is $90$ . Write a linear equation and find the numbers.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 8 core - answers

Fluency

Solve linear equations

$x = 6$ . Method: subtract $9$ from both sides.
$x = 6$ . Method: divide both sides by $3$ .
$x = 6$ . Method: subtract $5$ , then divide by $2$ .
$y = 7$ . Method: add $8$ , then divide by $3$ .
$x = 24$ . Method: subtract $3$ , then multiply by $4$ .
$a = 7$ . Method: divide both sides by $4$ to get $a - 2 = 5$ , then add $2$ .
$x = 5$ . Method: subtract $2x$ from both sides, then $3x - 3 = 12$ ; $3x = 15$ .
$y = 5$ . Method: subtract $3y$ and add $16$ : $20 = 4y$ , so $y = 5$ .
$x = 5$ . Method: multiply both sides by $2$ to get $3x + 1 = 16$ ; $3x = 15$ .
$x = 24$ . Method: multiply both sides by $12$ : $4x - 3x = 24$ .

Fluency

Solve inequalities

$x > 6$
$y \leq 4$
$x \geq 4$
$x > -5$ (flip: divide by $-4$ ).
$x < 2$ . Method: subtract $5$ , $-2x > -4$ ; divide by $-2$ , flip.
$y \geq 4$

Fluency

Tables and graphs

$y$ -values: $-5, -2, 1, 4, 7$ .
$y$ -intercept $(0, 7)$ .
Gradient $4$ .
$x$ -intercept $(3, 0)$ .
Gradient $3$ . Method: $\dfrac{9 - 3}{2 - 0}$ .
Yes: $2(3) - 1 = 5$ . ✓

Reasoning

Explain and spot the mistake

No. Dividing $12$ by $-3$ gives $-4$ , so $x = -4$ . Kai kept the sign wrong.
She forgot to flip the inequality when dividing by a negative. Correct: $-2x < 6$ gives $x > -3$ .
They both have gradient $2$ (same slope), but different $y$ -intercepts ( $0$ and $3$ ), so they rise equally but start at different heights.
The line shifts up by $c$ units on the $y$ -axis (parallel to itself). Increasing $c$ does not change the gradient.

Problem solving

Modelling

After $w = 11$ weeks (since $50 + 15 \times 10 = 200$ exactly at week $10$ ; pass $200 at end of week $11$ ). Strictly $50 + 15w > 200$ gives $w > 10$ , so first at $w = 11$ .
$V = 500 - 8t$ . Empty when $V = 0$ : $t = 62.5$ hours.
$16$ km. Method: $3.80 + 2k = 19.80$ , so $2k = 16$ .
$70$ texts. Method: $30 + 0.05t = 33.50$ , $0.05t = 3.50$ .
Plan A: $C = 20 + 0.15m$ ; Plan B: $C = 35 + 0.05m$ . Equal when $20 + 0.15m = 35 + 0.05m$ , so $0.10m = 15$ , $m = 150$ minutes.

Challenge - answers

Reasoning

Harder reasoning

$(5, 9)$ . Method: $2x - 1 = x + 4$ gives $x = 5$ , then $y = 9$ .
$x = 13$ . Method: multiply by $6$ : $3(x + 1) - 2(x - 1) = 12$ ; $3x + 3 - 2x + 2 = 12$ ; $x + 5 = 12$ .
$y = 3x + 1$ . Method: $y - 7 = 3(x - 2)$ , so $y = 3x - 6 + 7$ .
$28, 30, 32$ . Method: $n + (n+2) + (n+4) = 3n + 6 = 90$ ; $n = 28$ .

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