Year 7 Mathematics | Victorian Curriculum 2.0
Algebraic expressions
Topic 05 | Number & Algebra | Practice

Start here: letters stand for numbers we don’t know yet

You already use letters as placeholders. “Think of a number, double it, add three” — call the number nnn, and the instruction becomes 2n+32n + 32n+3. That’s algebra.

Three tricks that make everything easier:

  1. 3n3n3n means 3×n3 \times n3×n. We drop the ”×” because it looks like an xxx.
  2. Like terms can be added: 3x+2x=5x3x + 2x = 5x3x+2x=5x, because “three of something plus two of the same something equals five of that something”.
  3. Unlike terms cannot be combined: 3x+2y3x + 2y3x+2y stays as 3x+2y3x + 2y3x+2y — three xxxs and two yyys are different things.

What you will learn

  • use the language of algebra: term, coefficient, variable, constant, expression,
  • write an expression from a word description,
  • collect like terms,
  • substitute a value into an expression,
  • use everyday formulas (area, sporting scores, wages, density, heart rates) and substitute real values to find unknowns,
  • expand a single bracket: a(b+c)=ab+aca(b + c) = ab + aca(b+c)=ab+ac.
Why use letters instead of numbers?

Imagine explaining a game’s scoring rule: “multiply goals by six and add behinds.” That works for one sport. But 6g+b6g + b6g+b works for every game — you just plug in the numbers. Letters replace specific values so you can state a rule once and reuse it everywhere. That is the power of algebra: one expression handles infinitely many cases.

Where you'll see this
  • Formulas: A=L×WA = L \times WA=L×W (area), d=s×td = s \times td=s×t (distance), F=9/5 C+32F = 9/5 \, C + 32F=9/5C+32 (temperature).
  • Sport: AFL score = 6×goals+behinds6 \times \text{goals} + \text{behinds}6×goals+behinds.
  • Pay: weekly wage = hourly rate ×\times× hours worked.
  • Spreadsheets: every cell formula is algebra — the letter is the cell reference.
Worked example 0 Real-world example: AFL score formula

Collingwood kick 121212 goals and 888 behinds. What is their score?

The AFL formula is: Score =6g+b= 6g + b=6g+b, where ggg = goals and bbb = behinds.

  1. Substitute g=12g = 12g=12, b=8b = 8b=8: Score =6(12)+8=72+8=80= 6(12) + 8 = 72 + 8 = 80=6(12)+8=72+8=80.
  2. Their opponent scores 101010 goals, 141414 behinds: 6(10)+14=746(10) + 14 = 746(10)+14=74.
  3. Collingwood wins by 80−74=680 - 74 = 680−74=6 points.

Key idea: the formula 6g+b6g + b6g+b works for every game — you never rewrite the rule, just plug in new numbers. That is why algebra uses letters.

Try mentally first

Read 3x+53x + 53x+5 aloud as “three lots of xxx, plus five”. When you substitute a value, plug it in and do the arithmetic. If x=4x = 4x=4: three lots of 444, plus five — that’s 171717.

1. The language of algebra

A pronumeral (or variable) is a letter that stands for a number. In the expression 3x+53x + 53x+5:

  • 3x3x3x and 555 are terms (parts separated by +++ or −-−),
  • 333 is the coefficient of xxx (the number multiplying the pronumeral),
  • xxx is the variable,
  • 555 is a constant (it does not involve a variable).
Short-hand you can rely on

We write 3x3x3x rather than 3×x3 \times x3×x, and x5\dfrac{x}{5}5x​ rather than x÷5x \div 5x÷5. Unless brackets are used, 3x+53x + 53x+5 means (3×x)+5(3 \times x) + 5(3×x)+5.

Writing expressions from words

WordsExpression
555 more than nnnn+5n + 5n+5
333 less than nnnn−3n - 3n−3
Twice nnn2n2n2n
Half of nnnn2\dfrac{n}{2}2n​
Product of aaa and bbbababab
Sum of xxx and yyy, squared(x+y)2(x + y)^2(x+y)2

2. Collecting like terms

Like terms share the same variable part. For example, 3x3x3x and 5x5x5x are like terms; 3x3x3x and 3y3y3y are not. Only like terms can be added or subtracted.

Collecting like terms
3x+5x  =  8x,4y−y  =  3y,2a+5+3a−2  =  5a+3.3x + 5x \;=\; 8x, \qquad 4y - y \;=\; 3y, \qquad 2a + 5 + 3a - 2 \;=\; 5a + 3.3x+5x=8x,4y−y=3y,2a+5+3a−2=5a+3.
Worked example E Very easy: collect like terms

Simplify   2a+3a\;2a + 3a2a+3a.

Both are “lots of aaa”. Add the coefficients: 2+3=52 + 3 = 52+3=5.

2a+3a=5a.2a + 3a = 5a.2a+3a=5a.

Another:   7x−3x=4x\;7x - 3x = 4x7x−3x=4x. Same idea — four xxxs left over.

Worked example 1 Simplifying by collecting like terms

Simplify   7p+3q−2p+5q−4\;7p + 3q - 2p + 5q - 47p+3q−2p+5q−4.

  1. Put like terms next to each other:   (7p−2p)+(3q+5q)−4\;(7p - 2p) + (3q + 5q) - 4(7p−2p)+(3q+5q)−4.
  2. Combine: 5p+8q−45p + 8q - 45p+8q−4.

3. Substitution

Substitute means “replace the pronumeral with the given number, then evaluate”.

Worked example 2 Substitution

If a=3a = 3a=3 and b=−2b = -2b=−2, evaluate   4a−3b+1\;4a - 3b + 14a−3b+1.

4(3)−3(−2)+1  =  12+6+1  =  19.4(3) - 3(-2) + 1 \;=\; 12 + 6 + 1 \;=\; 19.4(3)−3(−2)+1=12+6+1=19.
Brackets save you

When substituting a negative, put the number in brackets. Without them, you will write 4−24-24−2 by accident instead of 4×(−2)4\times(-2)4×(−2).

Common mistake: combining unlike terms

3x+2y3x + 2y3x+2y is not 5xy5xy5xy. You can only collect terms that have the same variable part. 3x3x3x and 2y2y2y are different kinds of “thing” — like apples and bananas. Leave the expression as 3x+2y3x + 2y3x+2y.

4. Using everyday formulas

Algebra is how we write down real-world relationships compactly. Once you know a formula, you substitute values to find the unknown.

Formulas you will meet in Year 7

Rectangle
A=L×WandP=2L+2W.A = L \times W \qquad \text{and} \qquad P = 2L + 2W.A=L×WandP=2L+2W.

AAA is area, LLL is length, WWW is width, PPP is perimeter.

AFL score
total=6g+b.\text{total} = 6g + b.total=6g+b.

ggg is the number of goals (worth 666 points each), bbb is the number of behinds (111 point each).

Weekly wage with overtime
W=b+1.5×r×h.W = b + 1.5 \times r \times h.W=b+1.5×r×h.

bbb is the base wage for the week, rrr is the normal hourly rate, hhh is hours of overtime (paid at time-and-a-half).

Density
D=MV.D = \dfrac{M}{V}.D=VM​.

DDD is density, MMM is mass, VVV is volume. Common units: g/cm^3 or kg/m^3.

Maximum heart rate (rough estimate)
HRmax⁡≈220−a.HR_{\max} \approx 220 - a.HRmax​≈220−a.

aaa is age in years. A target zone for moderate exercise is about 50%50\%50% to 70%70\%70% of HRmax⁡HR_{\max}HRmax​.

Speed, distance, time
speed=distancetime,distance=speed×time.\text{speed} = \dfrac{\text{distance}}{\text{time}}, \quad \text{distance} = \text{speed} \times \text{time}.speed=timedistance​,distance=speed×time.
A formula is a recipe

Every letter stands for a real quantity. Three steps: (1) identify what each letter means, (2) substitute the numbers you know, (3) do the arithmetic.

Worked example F1 AFL score

Geelong kicked 121212 goals and 999 behinds. Using total=6g+b\text{total} = 6g + btotal=6g+b, find their score.

total=6×12+9=72+9=81 points.\text{total} = 6 \times 12 + 9 = 72 + 9 = 81 \text{ points}.total=6×12+9=72+9=81 points.
Worked example F2 Weekly wage with overtime

Lara has a base wage of $400 per week and a normal hourly rate of $25. Last week she did 666 hours of overtime. Using W=b+1.5rhW = b + 1.5rhW=b+1.5rh, find her total pay.

W=400+1.5×25×6=400+225=625.W = 400 + 1.5 \times 25 \times 6 = 400 + 225 = 625.W=400+1.5×25×6=400+225=625.

Her total pay is $625.

Worked example F3 Density

A wooden block has mass 270270270 g and volume 300300300 cm^3. Using D=MVD = \dfrac{M}{V}D=VM​, find its density.

D=270300=0.9 g/cm3.D = \frac{270}{300} = 0.9 \text{ g/cm}^3.D=300270​=0.9 g/cm3.

For reference, water has density 1.01.01.0 g/cm^3, so a lower number means the block will float.

Worked example F4 Target heart rate

Using HRmax⁡≈220−aHR_{\max} \approx 220 - aHRmax​≈220−a, find the maximum heart rate for a 131313-year-old, and the target upper bound at 70%70\%70%.

  1. Max: HRmax⁡=220−13=207HR_{\max} = 220 - 13 = 207HRmax​=220−13=207 beats per minute.
  2. Upper target: 70%70\%70% of 207=0.70×207≈145207 = 0.70 \times 207 \approx 145207=0.70×207≈145 bpm.

5. Expanding a single bracket

Expanding (sometimes called the distributive law) means multiplying every term inside the bracket by whatever is in front.

Distributive law
a(b+c)  =  ab+ac,a(b−c)  =  ab−ac.a(b + c) \;=\; ab + ac, \qquad a(b - c) \;=\; ab - ac.a(b+c)=ab+ac,a(b−c)=ab−ac.
Worked example 3 Expanding a bracket

Expand   3(2x+5)\;3(2x + 5)3(2x+5).

3(2x+5)  =  3×2x+3×5  =  6x+15.3(2x + 5) \;=\; 3 \times 2x + 3 \times 5 \;=\; 6x + 15.3(2x+5)=3×2x+3×5=6x+15.
Worked example 4 Expanding and collecting

Simplify   5(x+2)−3(x−4)\;5(x + 2) - 3(x - 4)5(x+2)−3(x−4).

  1. Expand each bracket: 5x+10−3x+125x + 10 - 3x + 125x+10−3x+12. Take care with the −3(x−4)=−3x+12-3(x - 4) = -3x + 12−3(x−4)=−3x+12.
  2. Collect like terms: (5x−3x)+(10+12)=2x+22(5x - 3x) + (10 + 12) = 2x + 22(5x−3x)+(10+12)=2x+22.

6. Simple algebraic fractions

6x+123  =  6x3+123  =  2x+4.\frac{6x + 12}{3} \;=\; \frac{6x}{3} + \frac{12}{3} \;=\; 2x + 4.36x+12​=36x​+312​=2x+4.

Every term in the numerator is divided by the denominator.


Practice

Fluency

Tier 1: basic skills

    1. Write an expression for “seven more than nnn”.
    2. Write an expression for “five less than 2m2m2m”.
    3. Write an expression for “the product of 444 and xxx”.
    4. Write an expression for “half of yyy added to 333”.
    5. In the term −7x-7x−7x, state the coefficient.
    6. Simplify   4x+3x\;4x + 3x4x+3x.
    7. Simplify   9y−5y\;9y - 5y9y−5y.
    8. Simplify   3a+2b+5a−b\;3a + 2b + 5a - b3a+2b+5a−b.
    9. Simplify   7p−4−2p+9\;7p - 4 - 2p + 97p−4−2p+9.
    10. Simplify   2x+5y+3x−y+4\;2x + 5y + 3x - y + 42x+5y+3x−y+4.
    11. Evaluate 3a+23a + 23a+2 when a=4a = 4a=4.
    12. Evaluate 5x−y5x - y5x−y when x=2x = 2x=2 and y=−3y = -3y=−3.
    13. Evaluate x2+1x^2 + 1x2+1 when x=−4x = -4x=−4.
    14. Expand   2(x+5)\;2(x + 5)2(x+5).
    15. Expand   4(y−3)\;4(y - 3)4(y−3).
    16. Expand   −3(a+2)\;-3(a + 2)−3(a+2).
    17. Expand   6(2m−1)\;6(2m - 1)6(2m−1).
    18. Simplify 8x+124\dfrac{8x + 12}{4}48x+12​.
    19. Simplify   3(x+2)+5\;3(x + 2) + 53(x+2)+5.
    20. Simplify   2(a+4)−3\;2(a + 4) - 32(a+4)−3.
Reasoning

Tier 2: mixed practice

    1. Simplify   4(x+3)+2(x−1)\;4(x + 3) + 2(x - 1)4(x+3)+2(x−1).
    2. Simplify   5(2y−1)−3(y+4)\;5(2y - 1) - 3(y + 4)5(2y−1)−3(y+4).
    3. Simplify   7m+4−2(m+3)\;7m + 4 - 2(m + 3)7m+4−2(m+3).
    4. Evaluate   3x2−2x+1\;3x^2 - 2x + 13x2−2x+1 when x=4x = 4x=4.
    5. Evaluate   (a+b)2\;(a + b)^2(a+b)2 when a=3a = 3a=3, b=−1b = -1b=−1.
    6. Expand   −2(3x−4)−(x+2)\;-2(3x - 4) - (x + 2)−2(3x−4)−(x+2).
    7. Simplify 10a−155\dfrac{10a - 15}{5}510a−15​.
    8. Write an expression for the perimeter of a rectangle with length 2x+12x + 12x+1 and width xxx. Simplify it.
    9. Write an expression for the cost of nnn apples at $0.60 each and mmm bananas at $0.40 each.
    10. Find the missing coefficient: 5x+□x=12x5x + \square x = 12x5x+□x=12x.
    11. Simplify   3a×4b\;3a \times 4b3a×4b.
    12. Simplify   12xy3y\;\dfrac{12xy}{3y}3y12xy​.
Reasoning

Tier 3: explain and spot the mistake

    1. Kira writes 3+2x=5x3 + 2x = 5x3+2x=5x. Is Kira correct? If not, explain the mistake and give the correct simplification.
    2. Explain why 3x3x3x and 3x23x^23x2 are not like terms, with a numerical example.
    3. Leo expands −4(x−3)-4(x - 3)−4(x−3) as −4x−12-4x - 12−4x−12. Is this right? If not, what is the correct expansion?
    4. Write two different expressions that both equal 121212 when x=3x = 3x=3.
    5. Are 2(a+3)2(a + 3)2(a+3) and 2a+32a + 32a+3 always equal? Explain with an example.
Fluency

Using everyday formulas (substitution)

    Use the formulas given in the Year 7 core section.

    1. A rectangle has L=12L = 12L=12 cm and W=5W = 5W=5 cm. Find its area and perimeter using A=L×WA = L \times WA=L×W and P=2L+2WP = 2L + 2WP=2L+2W.
    2. Hawthorn scored 141414 goals and 777 behinds. Use total=6g+b\text{total} = 6g + btotal=6g+b to find their total.
    3. Collingwood kicked 999 goals and 131313 behinds; Melbourne kicked 111111 goals and 555 behinds. Who won, and by how much?
    4. Sam’s base wage is $360 per week, normal rate $22/h. He worked 444 hours of overtime. Use W=b+1.5rhW = b + 1.5rhW=b+1.5rh to find his total pay.
    5. A metal bar has mass 504504504 g and volume 646464 cm^3. Find its density using D=MVD = \dfrac{M}{V}D=VM​.
    6. Find the maximum heart rate for a person aged 454545 using HRmax⁡=220−aHR_{\max} = 220 - aHRmax​=220−a.
    7. A car travels at 808080 km/h for 2.52.52.5 h. How far does it travel? (Use d=s×td = s \times td=s×t.)
    8. A cyclist covers 454545 km in 333 hours. Find the average speed.
    9. Use the formula C=59(F−32)C = \tfrac{5}{9}(F - 32)C=95​(F−32) to convert 68deg68 deg68degF to degrees Celsius.
Problem solving

Tier 4: real-world problems

    1. Mira has $x. She spends $5 on lunch and then earns $20 helping a neighbour. Write an expression for how much she has now. If she started with $12, how much has she now?
    2. A phone plan costs a $20 monthly fee plus $0.10 per minute. Write an expression for the cost of a month with ttt minutes of calls. What is the cost if t=150t = 150t=150?
    3. The length of a rectangle is 333 cm more than twice its width. If the width is www, write expressions for the length and perimeter. Simplify the perimeter.
    4. A taxi charges a $4.50 flag-fall plus $2 per kilometre. Write the cost for a kkk-kilometre trip, and find the cost of a 121212 km trip.
    5. Five students each give $x toward a gift that costs $42. Write an expression for how much change is left after the gift is bought. Evaluate it if x=10x = 10x=10.
    6. A swimming pool holds VVV litres and a hose fills it at rrr L/min. Write an expression for the time to fill the pool. How long (in minutes) if V=60 000V = 60\,000V=60000 and r=150r = 150r=150?
    7. A mobile plan charges $25 per month plus $0.08 per text. Lucy sent ttt texts in a month. Write an expression for her total cost, then find the cost when t=180t = 180t=180.
    8. A gym membership costs $59 to join plus $15 per week. Write an expression for the total cost after www weeks. When does the total first exceed $200?
    9. A delivery van’s fuel cost per trip is C=0.15dC = 0.15 dC=0.15d, where ddd is the trip distance in kilometres. Find the cost of a 240240240 km trip. If the fuel cost doubled per kilometre, what would the new formula be?
    10. Daniella’s target training heart rate zone is between 50%50\%50% and 70%70\%70% of HRmax⁡=220−aHR_{\max} = 220 - aHRmax​=220−a (where aaa is her age in years). She is 121212. Find the two ends of her target zone.
Year 7 Mathematics study companion | Practice