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YEAR 10 · ALGEBRA · 15 MIN

Simplifying expressions

Collect like terms and use the structure of an expression to simplify it without changing its value.

In this lesson

Before you start
  • Multiplying signed numbers
  • Variables and powers
01 / LEARN

Understand the idea.

Which terms can be combined?

Like terms have exactly the same variable part, including every power. In 7⁢a−3⁢b, the coefficients are 7 and −3: the minus belongs to the second term. Collecting like terms changes their coefficients, while leaving their variable part unchanged.

Simplifying in a reliable order

Expand any brackets first, then organise terms by variable and power. A subtraction before a bracket affects every term inside it. Keep constants in their own group so that none are accidentally combined with variable terms.

Addition and multiplication behave differently: 3⁢x+2⁢x=5⁢x, whereas 3⁢x×2⁢x=6x2. Substituting a number into both expressions can expose an error, but agreement at one input alone does not prove they are identical.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Collect coefficients

7⁢a−3⁢b−2⁢a+5⁢b

Worked solution

  1. 1

    Group terms with matching variables.

    7⁢a−3⁢b−2⁢a+5⁢b=(7⁢a−2⁢a)+(−3⁢b+5⁢b)

    Reordering addition is valid when each sign stays with its term.

  2. 2

    Add the coefficients in each group.

    7⁢a−2⁢a=(7−2)⁢a=5⁢a−3⁢b+5⁢b=(−3+5)⁢b=2⁢b

    Each group counts multiples of the same variable.

  3. 3

    Write the simplified expression.

    5⁢a+2⁢b

    The remaining variable parts differ, so these two terms cannot be collected further.

Answer

5⁢a+2⁢b

EXAMPLE 02

Keep powers distinct

Simplify 4x2−3⁢x+6−2x2+5⁢x−9.

Worked solution

  1. 1

    Separate square terms, linear terms and constants.

    (4x2−2x2)+(−3⁢x+5⁢x)+(6−9)

    The powers of the variable must match before terms can be combined.

  2. 2

    Combine each group.

    (4−2)x2=2x2(−3+5)⁢x=2⁢x6−9=−3

    Only the coefficients change when like terms are added.

  3. 3

    Assemble the result in descending powers.

    2x2+2⁢x−3

    The square term, linear term and constant remain separate.

Answer

2x2+2⁢x−3

03 / YOUR TURN

Try it yourself.

Check your understanding

Simplify 6⁢p−2⁢q−9⁢p+5⁢q.

Show answer & reasoning

−3⁢p+3⁢q

Collect the p terms to get (6−9)⁢p=−3⁢p and the q terms to get (−2+5)⁢q=3⁢q. Therefore the expression is −3⁢p+3⁢q.

Keep practising

6 worksheet questions · 14 marks

Read the questions online
  1. Simplify 9⁢x−4⁢x+2⁢x.
    2 marks
  2. Simplify 5⁢a+3⁢b−2⁢a+4⁢b.
    2 marks
  3. Simplify 8m2−3⁢m+2m2+7⁢m.
    2 marks
  4. Simplify 4⁢p−(3⁢p−5).
    2 marks
  5. A rectangle has adjacent sides 3⁢x+2 cm and x+5 cm. Write its perimeter in simplified form.
    3 marks
  6. Rina says 2⁢x+5⁢y=7⁢x⁢y. Explain the error and test the claim using x=2,y=1.
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Simplify 9⁢x−4⁢x+2⁢x.
    2 marks

    Add coefficients: 9−4+2=7.

    7⁢x
  2. Simplify 5⁢a+3⁢b−2⁢a+4⁢b.
    2 marks

    Collect 5⁢a−2⁢a and 3⁢b+4⁢b.

    3⁢a+7⁢b
  3. Simplify 8m2−3⁢m+2m2+7⁢m.
    2 marks

    Collect equal powers separately.

    10m2+4⁢m
  4. Simplify 4⁢p−(3⁢p−5).
    2 marks

    The subtraction gives 4⁢p−3⁢p+5.

    p+5
  5. A rectangle has adjacent sides 3⁢x+2 cm and x+5 cm. Write its perimeter in simplified form.
    3 marks

    P=2⁢(3⁢x+2)+2⁢(x+5).

    Expand and collect: 6⁢x+4+2⁢x+10=8⁢x+14.

    8⁢x+14 cm
  6. Rina says 2⁢x+5⁢y=7⁢x⁢y. Explain the error and test the claim using x=2,y=1.
    3 marks

    The variable parts x and y differ.

    Substitution gives 2⁢(2)+5⁢(1)=9, while 7⁢(2)⁢(1)=14.

    Unlike terms cannot be combined. The original equals 9; 7⁢x⁢y equals 14.
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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 algebra, including the approved fuller school-course treatment of rational expressions and formula restrictions; quadratic methods are in their own chapter.

Official curriculum source ↗

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