Year 8 Mathematics | Victorian Curriculum 2.0
Linear equations, inequalities & graphs
Topic 07 | Number & Algebra | Practice

Start here: two ideas, one flow

This topic puts three connected ideas on one page:

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

Try mentally first

For equations: “what do I do to both sides to get xxx alone?” For graphs: “what is yyy when x=0x = 0x=0, and by how much does yyy change when xxx goes up by 111?” Those two questions unlock most of this topic.

What you will learn

  • solve linear equations of the form ax+b=cax + b = cax+b=c and ax+b=cx+dax + b = cx + dax+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+cy = mx + cy=mx+c,
  • read and interpret the yyy-intercept ccc (starting value) and gradient mmm (rate of change),
  • use linear relations to model practical problems (taxi fares, pay rates, tank drainage).
Where you'll see this
  • Taxi fares: flag-fall plus per-km rate is a linear model.
  • Phone plans: fixed monthly fee plus per-minute rate.
  • Savings: balance = starting amount + weekly deposit ×\times× weeks.
  • Physics: distance-time graphs at constant speed are straight lines.

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\;x + 7 = 12x+7=12.

The left-hand side says ”xxx plus 777”. To get xxx alone, undo the “plus 777” by subtracting 777 from both sides:

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

Check: 5+7=125 + 7 = 125+7=12. ✓

Worked example 1 Pronumeral on one side

Solve   4x−5=23\;4x - 5 = 234x−5=23.

  1. Add 555: 4x=284x = 284x=28.
  2. Divide by 444: x=7x = 7x=7.

Check: 4(7)−5=28−5=234(7) - 5 = 28 - 5 = 234(7)−5=28−5=23. ✓

Worked example 2 Pronumeral on both sides

Solve   5x+4=2x+19\;5x + 4 = 2x + 195x+4=2x+19.

  1. Subtract 2x2x2x from both sides: 3x+4=193x + 4 = 193x+4=19.
  2. Subtract 444: 3x=153x = 153x=15.
  3. Divide by 333: x=5x = 5x=5.

Check: 5(5)+4=295(5) + 4 = 295(5)+4=29 and 2(5)+19=292(5) + 19 = 292(5)+19=29. ✓

Worked example 3 With brackets and fractions

Solve   2x−13=5\;\dfrac{2x - 1}{3} = 532x−1​=5.

Multiply both sides by 333: 2x−1=152x - 1 = 152x−1=15. Then 2x=162x = 162x=16, x=8x = 8x=8.

2. Linear inequalities

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

Flip when multiplying or dividing by a negative

Multiplying or dividing both sides of an inequality by a negative number flips the direction of the inequality. −2x≤6-2x \leq 6−2x≤6 gives x≥−3x \geq -3x≥−3 (not x≤−3x \leq -3x≤−3).

Common mistake: doing to one side only

Whatever you do to one side of an equation must be done to the other side too, or the scales tip and the equation stops being true. If you subtract 333 from the left, you must subtract 333 from the right as well.

Worked example E2 Very easy: inequality in one step

Solve   x+3>10\;x + 3 > 10x+3>10.

Subtract 333 from both sides: x>7x > 7x>7. No flip needed because we did not multiply or divide by a negative.

Worked example 4 Solve an inequality

Solve   3x−4≤11\;3x - 4 \leq 113x−4≤11.

  1. Add 444: 3x≤153x \leq 153x≤15.
  2. Divide by 333 (positive, no flip): x≤5x \leq 5x≤5.

On a number line, draw a filled circle at 555 and shade to the left.

Try this first

Before moving on, check: can you solve 2x+1=92x + 1 = 92x+1=9 in your head? If you got x=4x = 4x=4, you are ready for graphs.

3. Linear relations on the Cartesian plane

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

Reading a linear graph

Standard form
y=mx+c.y = mx + c.y=mx+c.
  • ccc is the yyy-intercept: where the line crosses the yyy-axis (the value of yyy when x=0x = 0x=0).
  • mmm is the gradient (slope): the change in yyy per unit change in xxx.
Table of values

Pick a few xxx-values, compute yyy, 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 yyy-intercept of y=3x+5y = 3x + 5y=3x+5?

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

The gradient is the number in front of x, not the constant

In y=2x+3y = 2x + 3y=2x+3, the gradient is 222 (not 333). The constant 333 is the yyy-intercept. A common mix-up is to read the last number as the gradient.

Worked example 5 Plotting y = 2x - 1

Table of values:

xxx−1-1−1000111222333
yyy−3-3−3−1-1−1111333555
  • yyy-intercept: (0,−1)(0, -1)(0,−1) - the line crosses the yyy-axis at −1-1−1.
  • Gradient: 222 - for every 111 unit right, yyy rises by 222.
xy3-23-3(0, -1)run 1rise 2y = 2x - 1
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.
xyy = x + 1y = 2x - 1y = -x + 4
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>0m > 0m>0) tilts upward left-to-right; a negative gradient (m<0m < 0m<0) tilts downward. A larger ∣m∣|m|∣m∣ makes the line steeper.

Gradient is rise over run, not run over rise

If a line goes up 444 units for every 222 units across, the gradient is 42=2\dfrac{4}{2} = 224​=2, not 24\dfrac{2}{4}42​. Always put the vertical change on top.

Try this first

Look at y=−x+4y = -x + 4y=−x+4 in the diagram above. Can you identify its gradient and yyy-intercept? (Answer: m=−1m = -1m=−1, c=4c = 4c=4.) If so, move on.

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+2.5kF = 4 + 2.5kF=4+2.5k where kkk is the distance in km.
  • yyy-intercept (flag-fall): $4.
  • Gradient: $2.50 per km.
  • A 121212 km trip costs 4+2.5×12=4+30=344 + 2.5 \times 12 = 4 + 30 = 344+2.5×12=4+30=34 dollars.

From a graph of FFF against kkk, a $20 fare corresponds to where the line hits F=20F = 20F=20, i.e. 4+2.5k=204 + 2.5k = 204+2.5k=20, so k=6.4k = 6.4k=6.4 km.


Practice: Year 8 core

Fluency

Solve linear equations

    1. Solve x+9=15x + 9 = 15x+9=15.
    2. Solve 3x=183x = 183x=18.
    3. Solve 2x+5=172x + 5 = 172x+5=17.
    4. Solve 3y−8=133y - 8 = 133y−8=13.
    5. Solve x4+3=9\dfrac{x}{4} + 3 = 94x​+3=9.
    6. Solve 4(a−2)=204(a - 2) = 204(a−2)=20.
    7. Solve 5x−3=2x+125x - 3 = 2x + 125x−3=2x+12.
    8. Solve 3y+4=7y−163y + 4 = 7y - 163y+4=7y−16.
    9. Solve 3x+12=8\dfrac{3x + 1}{2} = 823x+1​=8.
    10. Solve x3−x4=2\dfrac{x}{3} - \dfrac{x}{4} = 23x​−4x​=2.
Fluency

Solve inequalities

    1. Solve x+4>10x + 4 > 10x+4>10.
    2. Solve 2y−3≤52y - 3 \leq 52y−3≤5.
    3. Solve 3x+1≥133x + 1 \geq 133x+1≥13.
    4. Solve −4x<20-4x < 20−4x<20.
    5. Solve 5−2x>15 - 2x > 15−2x>1.
    6. Solve y2+1≥3\dfrac{y}{2} + 1 \geq 32y​+1≥3.
Fluency

Tables and graphs

    1. Complete the table for y=3x−2y = 3x - 2y=3x−2:

      xxx−1-1−1000111222333
      yyy?????
    2. Find the yyy-intercept of y=−2x+7y = -2x + 7y=−2x+7.

    3. Find the gradient of y=4x−1y = 4x - 1y=4x−1.

    4. Find the xxx-intercept of y=2x−6y = 2x - 6y=2x−6 (where y=0y = 0y=0).

    5. A line passes through (0,3)(0, 3)(0,3) and (2,9)(2, 9)(2,9). Find the gradient.

    6. Does the point (3,5)(3, 5)(3,5) lie on y=2x−1y = 2x - 1y=2x−1? Show by substitution.

Reasoning

Explain and spot the mistake

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

Modelling

    1. A gym charges a $50 joining fee and $15 per week. Write a formula C=50+15wC = 50 + 15wC=50+15w for cost after www weeks. After how many weeks does the total pass $200?
    2. A water tank starts with 500500500 L and leaks at 888 L per hour. Write a formula for the volume VVV after ttt hours. When is the tank empty?
    3. A taxi has flag-fall $3.80 and charges $2 per km. A trip costs $19.80. How long was it (in km)?
    4. 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?
    5. 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

    1. Solve simultaneously (by substitution): y=2x−1y = 2x - 1y=2x−1 and y=x+4y = x + 4y=x+4.
    2. Solve x+12−x−13=2\dfrac{x + 1}{2} - \dfrac{x - 1}{3} = 22x+1​−3x−1​=2.
    3. A line has gradient 333 and passes through (2,7)(2, 7)(2,7). Find the equation of the line.
    4. The sum of three consecutive even numbers is 909090. Write a linear equation and find the numbers.
Year 8 Mathematics study companion | Practice