FA2, FA5, HA2, HA5 - Substituting into formulae

FA2, FA5, HA2, HA5 - Substituting into formulae

One formula can describe many different rectangles, journeys or experiments. Learn to replace its letters with values, keep negative signs and powers in the right places, and interpret the result with suitable units. FA2 and HA2 cover substitution, including scientific formulae; FA5 and HA5 contribute the use of standard mathematical formulae.

Positive substitutions

Can a rectangle become longer without its perimeter doubling? A formula lets us calculate what actually changes instead of guessing from one length.

Substitution means replacing every occurrence of a letter by its given value. If a=4a=4 and b=5b=5, then 3a+2b=3×4+2×5=223a+2b=3\times4+2\times5=22. The letter aa does not mean 11, and 3a3a means multiplication, not the two-digit number 3434.

Formula

A formula is an equation connecting quantities. In P=2(l+w)P=2(l+w), PP is a rectangle's perimeter, ll its length and ww its width, all measured in compatible length units.

A rectangle has two lengths and two widths, so P=l+w+l+w=2(l+w)P=l+w+l+w=2(l+w). Write the substituted calculation before evaluating it. Brackets, powers, multiplication and division keep their usual priority after letters have been replaced.

Worked example 1: whole-number lengths

Question. Use P=2(l+w)P=2(l+w) to find the perimeter when l=8 cml=8\,\mathrm{cm} and w=3 cmw=3\,\mathrm{cm}.

Plan. Replace both dimensions, add inside the bracket, then double the result.

P=2(8+3)substitute both values=2×11add inside the bracket=22multiply by two\begin{aligned} P&=2(8+3) &&\text{substitute both values}\\ &=2\times11 &&\text{add inside the bracket}\\ &=22 &&\text{multiply by two} \end{aligned}

Answer. The perimeter is 22 cm22\,\mathrm{cm}.

Check. Adding all four sides gives 8+3+8+3=22 cm8+3+8+3=22\,\mathrm{cm}, agreeing with the formula.

Now only the width changes, to a decimal. The formula and the order of operations stay the same.

Worked example 2: a decimal length

Question. Use P=2(l+w)P=2(l+w) when l=8 cml=8\,\mathrm{cm} and w=3.5 cmw=3.5\,\mathrm{cm}.

Plan. Substitute the new width and complete the bracket before multiplying.

P=2(8+3.5)replace each letter=2×11.5evaluate the bracket=23double the sum\begin{aligned} P&=2(8+3.5) &&\text{replace each letter}\\ &=2\times11.5 &&\text{evaluate the bracket}\\ &=23 &&\text{double the sum} \end{aligned}

Answer. The perimeter is 23 cm23\,\mathrm{cm}.

Check. Each of the two widths increased by 0.5 cm0.5\,\mathrm{cm}, so the perimeter increased by 1 cm1\,\mathrm{cm}, from 2222 to 23 cm23\,\mathrm{cm}.

The number immediately before a bracket multiplies the whole bracket.

Watch out

2(8+3.5)≠2×8+3.52(8+3.5)\ne 2\times8+3.5. The second expression doubles only the length. Either add the bracket first or expand correctly to 2×8+2×3.52\times8+2\times3.5.

Use the same perimeter formula with a fresh pair of lengths.

Signed substitutions

A negative value must replace the whole letter, including its sign.

Suppose x=−4x=-4. Writing x2=(−4)2x^2=(-4)^2 shows that both factors are negative: (−4)×(−4)=16(-4)\times(-4)=16. By contrast, −42=−(4×4)=−16-4^2=-(4\times4)=-16: the leading minus is outside the power. Brackets make this difference visible.

A substituted negative value

When replacing a letter by a negative number, put that number in brackets: 3x=3(−4)3x=3(-4), x2=(−4)2x^2=(-4)^2 and 5−x=5−(−4)5-x=5-(-4) when x=−4x=-4.

Evaluate powers before products, then add or subtract. In particular, subtracting a negative is addition.

Worked example 3: a positive input

Question. Work out x2−3xx^2-3x when x=4x=4.

Plan. Replace both copies of xx, then evaluate the square and the product separately.

x2−3x=42−3×4substitute consistently=16−12power and product first=4subtract\begin{aligned} x^2-3x&=4^2-3\times4 &&\text{substitute consistently}\\ &=16-12 &&\text{power and product first}\\ &=4 &&\text{subtract} \end{aligned}

Answer. 44.

Check. x2−3x=x(x−3)x^2-3x=x(x-3); using x=4x=4 gives 4(4−3)=44(4-3)=4.

Change just the sign of the input. The squared term stays positive, but the product 3x3x changes sign.

Worked example 4: a negative input

Question. Work out x2−3xx^2-3x when x=−4x=-4.

Plan. Bracket each negative replacement before doing any arithmetic.

x2−3x=(−4)2−3(−4)include both minus signs=16−(−12)evaluate square and product=28subtracting a negative adds\begin{aligned} x^2-3x&=(-4)^2-3(-4) &&\text{include both minus signs}\\ &=16-(-12) &&\text{evaluate square and product}\\ &=28 &&\text{subtracting a negative adds} \end{aligned}

Answer. 2828.

Check. The equivalent form gives (−4)(−4−3)=(−4)(−7)=28(-4)(-4-3)=(-4)(-7)=28.

The exponent belongs to the expression immediately before it.

Watch out

If x=−4x=-4, then x2=16x^2=16, while −x2=−16-x^2=-16. Replacing xx by −4-4 does not remove the separate minus in front of x2x^2.

Keep the negative input visible until the individual terms are evaluated.

Scientific formulae

The units attached to a formula tell you which numerical values to substitute.

For a journey, average speed is total distance divided by total time. Using metres and seconds gives metres per second. If a formula specifies seconds, a time in minutes must first be converted; otherwise the calculation answers a different question.

The scientific formula E=12mv2E=\frac12mv^2 connects kinetic energy EE, mass mm and speed vv. For this formula, mm is measured in kilograms and vv in metres per second; the resulting energy is in joules, J\mathrm J. The square applies to vv, not to the whole product. This formula is supplied in the examples below; it is not on the Eduqas mathematical formula lists. Formula provision does not replace reading its quantities and units.

Worked example 5: compatible units

Question. Use E=12mv2E=\frac12mv^2 to calculate the kinetic energy of a 2 kg2\,\mathrm{kg} object moving at 3 m/s3\,\mathrm{m/s}.

Plan. Check the units, square the speed, then multiply by half the mass.

E=12×2×32substitute in the required units=12×2×9square the speed first=9multiply the factors\begin{aligned} E&=\tfrac12\times2\times3^2 &&\text{substitute in the required units}\\ &=\tfrac12\times2\times9 &&\text{square the speed first}\\ &=9 &&\text{multiply the factors} \end{aligned}

Answer. 9 J9\,\mathrm J.

Check. Half the mass is 1 kg1\,\mathrm{kg}, so the numerical energy is 1×9=91\times9=9, not (12×2×3)2(\tfrac12\times2\times3)^2.

Now change only the way the speed is given. A centimetre is one hundredth of a metre, so divide a speed in centimetres per second by 100100 to express it in metres per second.

Worked example 6: convert before substituting

Question. Use E=12mv2E=\frac12mv^2 for a 2 kg2\,\mathrm{kg} object moving at 300 cm/s300\,\mathrm{cm/s}. The formula requires vv in m/s\mathrm{m/s}.

Plan. Convert the speed first, then use the same energy calculation.

v=300÷100=3 m/scentimetres to metresE=12×2×32substitute the converted value=9 Jevaluate\begin{aligned} v&=300\div100=3\,\mathrm{m/s} &&\text{centimetres to metres}\\ E&=\tfrac12\times2\times3^2 &&\text{substitute the converted value}\\ &=9\,\mathrm J &&\text{evaluate} \end{aligned}

Answer. 9 J9\,\mathrm J.

Check. 300 cm300\,\mathrm{cm} and 3 m3\,\mathrm m describe the same distance. The physical speed has not changed, so neither should the energy.

A formula can describe how a result changes too. In the opening rectangle, doubling only ll gives 2(2l+w)2(2l+w), which generally differs from twice 2(l+w)2(l+w), because the width did not double.

Exam tip

Write the numerical substitution with its multiplication signs and brackets. It shows how you used the formula and makes a calculator or unit error easier to locate.

Convert the stated unit before evaluating this new case.

Quick Recap

  • Replace every occurrence of a letter by the same given value.
  • Keep the original operations and their priority.
  • Bracket negative replacements, especially before powers.
  • Convert values to the units required by a formula and label the result.

Doubling a rectangle's length alone does not generally double its perimeter: the two widths still contribute the same amount.