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YEAR 11 · LINEAR AND EXPONENTIAL MODELS · 18 MIN

Linear equations and formulas

Solve linear equations by balancing both sides, then rearrange and use practical formulas.

In this lesson

Before you start
  • Substituting into a formula
  • Simple equation solving
01 / LEARN

Understand the idea.

Keep the equation balanced

An equation makes a claim of equality. Keep that claim true by applying the same operation to both sides. To isolate a variable, undo addition or subtraction before undoing the multiplication around it.

A solution must satisfy the original equation. Substitution checks both your arithmetic and your rearrangement: calculate the original left side using your proposed value and compare it with the right side.

3⁢x+7=25

Choose the subject of a formula

In a practical formula, identify the unknown and attach units to the known quantities before calculating. In C=F+r⁢d, the fixed fee F is paid once; the term r⁢d depends on distance.

To recover distance, remove the fixed fee before dividing by the rate: d=C−Fr. This requires r≠0. The numerator is the distance-dependent charge, so dividing dollars by dollars per kilometre produces kilometres.

C=F+r⁢d⟹d=C−Fr,r≠0
02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Solve and check

Solve 3⁢x+7=25.

Worked solution

  1. 1

    Remove the added seven.

    3⁢x=25−7=18

    Subtracting seven from both sides preserves equality.

  2. 2

    Undo multiplication by three.

    x=183=6

    Every term on both sides is divided by the same nonzero number.

  3. 3

    Check in the original equation.

    3⁢(6)+7=18+7=25

    The proposed value makes the original sides equal.

Answer

x=6

EXAMPLE 02

Reverse a delivery rule

A delivery costs C=4+2.5⁢d dollars for a distance of d km. A bill is 24 dollars. Find the distance.

Worked solution

  1. 1

    Substitute the bill amount.

    24=4+2.5⁢d

    The formula gives the total charge, including the fixed fee.

  2. 2

    Remove the fixed fee.

    2.5⁢d=24−4=20

    Only the remaining twenty dollars pays for distance.

  3. 3

    Divide by the distance rate.

    d=202.5=8 km

    The rate is 2.50 dollars for each kilometre.

  4. 4

    Verify the delivery charge.

    4+2.5⁢(8)=24

    Eight kilometres reproduces the given bill.

Answer

8 km

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve 5⁢x−4=31.

Show answer & reasoning

x=7.

Add four to both sides to obtain 5⁢x=35, then divide by five: x=7. Substitution gives 5⁢(7)−4=31.

Keep practising

6 worksheet questions · 12 marks

Read the questions online
  1. Solve 4⁢x+9=37.
    2 marks
  2. Solve 7−2⁢x=−5.
    2 marks
  3. Solve 3⁢(x+2)=21 and check.
    2 marks
  4. Make h the subject of A=b⁢h2, where b≠0.
    2 marks
  5. Use C=18+6⁢n to find the cost for five items.
    2 marks
  6. A tank has V=150−12⁢t litres after t minutes. When does it hold 90 L?
    2 marks
Read the worked solutions

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

  1. Solve 4⁢x+9=37.
    2 marks

    Subtract 9: 4⁢x=28.

    Divide by 4.

    x=7.
  2. Solve 7−2⁢x=−5.
    2 marks

    Subtract 7: −2⁢x=−12.

    Divide by −2.

    x=6.
  3. Solve 3⁢(x+2)=21 and check.
    2 marks

    Divide by 3: x+2=7.

    Subtract 2; 3⁢(5+2)=21.

    x=5.
  4. Make h the subject of A=b⁢h2, where b≠0.
    2 marks

    Multiply both sides by 2.

    Divide the entire expression 2⁢A by b.

    h=2⁢Ab.
  5. Use C=18+6⁢n to find the cost for five items.
    2 marks

    C=18+6⁢(5)=48.

    48 dollars.
  6. A tank has V=150−12⁢t litres after t minutes. When does it hold 90 L?
    2 marks

    90=150−12⁢t gives 12⁢t=60.

    Thus t=5.

    5 min.
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

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