2.3 - Forming expressions and formulae

2.3 - Forming expressions and formulae

Turn verbal rules, labelled dimensions and practical situations into algebra. You will define variables, choose operations and derive expressions and formulae. The same rule can then answer many questions, from a rectangle's area to which printing service costs less.

Translating words into symbols

A printing service charges £3 plus £2 per poster. Another charges £7 plus £1.50 per poster. Which is cheaper for ten posters? The lower fixed fee alone does not settle it. We will turn each price rule into a formula and answer this by the end.

You already know

Substitution replaces a letter by its value. Brackets preserve order: “add 33, then double” gives 2(x+3)2(x+3); “double, then add 33” gives 2x+32x+3.

First say what each letter measures, including its unit. “Per kilometre” means multiply the rate by distance measured in kilometres. “Plus a fixed charge” means add that amount once.

Expression and formula

An expression, such as 4+1.5d4+1.5d, represents a quantity. An equation states that two expressions are equal. A formula is an equation expressing a rule between named quantities, such as C=4+1.5dC=4+1.5d. An expression alone has no equality.

A formula lets the input vary. A numerical equation such as 4+1.5d=134+1.5d=13 instead asks which distance gives a specified cost. There is no general words-to-symbols formula on either the Foundation or Higher formulae sheet: learn to translate the meaning.

Worked example 1: a fixed charge and a rate

Question. A taxi charges £4 plus £1.50 per kilometre. Write a formula for cost CC pounds for distance dd kilometres. Find the cost when d=6d=6.

Plan. Add the fixed charge to the rate multiplied by distance, then substitute.

distance charge=1.5drate times distanceC=4+1.5dadd the fixed charge onceC=4+1.5×6=13substitute d=6\begin{aligned} \text{distance charge}&=1.5d &&\text{rate times distance}\\ C&=4+1.5d &&\text{add the fixed charge once}\\ C&=4+1.5\times6=13 &&\text{substitute }d=6 \end{aligned}

Why multiply only the rate by dd? The fixed charge is paid once, regardless of distance.

Answer. C=4+1.5dC=4+1.5d; the journey costs £13.00.

Check. At d=0d=0, the formula gives the fixed £4. Each extra kilometre adds £1.50.

Change only the fixed charge from £4 to £6. The distance term stays unchanged.

Worked example 2: change the fixed contribution

Question. A taxi charges £6 plus £1.50 per kilometre. Write a formula for CC pounds for dd kilometres, and find CC when d=6d=6.

Plan. Replace the fixed term while keeping the rate term.

C=6+1.5dcombine the two contributionsC=6+1.5×6substitute the distance=15add 6+9\begin{aligned} C&=6+1.5d &&\text{combine the two contributions}\\ C&=6+1.5\times6 &&\text{substitute the distance}\\ &=15 &&\text{add }6+9 \end{aligned}

Why has the cost risen by exactly £2? Only the fixed charge changed, by £2.

Answer. C=6+1.5dC=6+1.5d; the cost is £15.00.

Check. The distance charge remains £9, and £9 plus the £6 fee is £15.

These models use d≥0d\ge0. Fractional distances are allowed; the domain follows the meaning of distance.

Watch out

Wrong: C=(6+1.5)dC=(6+1.5)d. It multiplies the fixed fee by every kilometre. Correct: C=6+1.5dC=6+1.5d. At d=0d=0, the wrong formula gives zero instead of the fixed fee.

Now change only the rate, and keep the fixed charge separate.

Deriving expressions from dimensions

A labelled diagram supplies relationships as well as numbers. Perimeter adds boundary lengths; area measures the space inside. Identify the required quantity before choosing an operation, and use the labels rather than measuring a picture.

[DIAGRAM: asset_slug: m030_formula_rectangle; caption: A rectangle with length x plus 4 centimetres and width x plus 1 centimetres. The diagram is not to scale; description: A rectangle has horizontal side labelled (x+4) cm and vertical side labelled (x+1) cm. Opposite sides are equal and each angle is a right angle. Use the labels, not measured pixel lengths; file: m030_formula_rectangle.png; recommended_method: matplotlib]
A rectangle with length x plus 4 centimetres and width x plus 1 centimetres. The diagram is not to scale

Rectangle perimeter and area

P=2l+2w,A=lwP=2l+2w,\qquad A=lw

For positive length ll and width ww, perimeter uses length units and area uses squared units. Foundation and Higher: Not in the formulae sheet: learn it.

Keep each multi-term dimension in brackets until you have multiplied it correctly.

Worked example 3: formulae from the rectangle

Question. A rectangle has length (x+4) cm(x+4)\,\mathrm{cm} and width (x+1) cm(x+1)\,\mathrm{cm}, where x>−1x>-1. Derive simplified formulae for perimeter PP in centimetres and area AA in square centimetres. Check at x=2x=2.

Plan. Add two of each side for perimeter and multiply the dimensions for area.

P=2(x+4)+2(x+1)four boundary sides=4x+10expand and collectA=(x+4)(x+1)length times width=x2+5x+4expand the product\begin{aligned} P&=2(x+4)+2(x+1) &&\text{four boundary sides}\\ &=4x+10 &&\text{expand and collect}\\ A&=(x+4)(x+1) &&\text{length times width}\\ &=x^2+5x+4 &&\text{expand the product} \end{aligned}

Why are the units different? Perimeter adds lengths; area multiplies two lengths.

Answer. P=4x+10P=4x+10 and A=x2+5x+4A=x^2+5x+4, in the stated units.

Check. At x=2x=2, the sides are 6 cm6\,\mathrm{cm} and 3 cm3\,\mathrm{cm}. Both formulae give the expected values: P=18 cmP=18\,\mathrm{cm}, A=18 cm2A=18\,\mathrm{cm}^2. Equal numerical values do not make these the same quantity.

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