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Rearranging formulas

Algebra · level 5, lesson 3 of 6

A question to start

A weather report from the United States says it will be 86 degrees Fahrenheit today. Is that cold, or hot?

You may know the formula that turns Celsius into Fahrenheit:

\[F = \frac{9}{5}C + 32.\]

But here you know \(F\) and want \(C\). You could put \(F = 86\) in and solve an equation: \(86 = \tfrac95 C + 32\). That works, but tomorrow the forecast will say 59 degrees and you will have to solve a new equation, and again the day after. It would be far better to turn the formula around once, so that it starts "\(C = \ldots\)". Then every conversion is a single calculation.

Turning a formula around is called rearranging, or changing the subject. By the end of this lesson you will be able to do it for any formula you meet in school science. (86 °F is 30 °C, a hot day; see worked example 2.)

The idea

The subject of a formula

A formula is an equation that connects several quantities, each with its own letter. The letter standing alone on one side is the formula's subject. In \(F = \tfrac95 C + 32\) the subject is \(F\): the formula tells you how to work out \(F\) when you know \(C\).

To make \(C\) the subject means to rewrite the formula, without changing what it says, so that \(C\) stands alone on one side.

Rearranging is solving, with letters

Here is the key idea of the lesson. Rearranging is exactly the same as solving an equation. The only difference is that some of the other quantities are letters instead of numbers. Every move you know still works, for the same reason: whatever you do to one side, you do to the other, and the balance stays level.

Compare these two side by side:

Solving \(86 = \tfrac95 C + 32\) Rearranging \(F = \tfrac95 C + 32\)
subtract 32: \(54 = \tfrac95 C\) subtract 32: \(F - 32 = \tfrac95 C\)
multiply by 5: \(270 = 9C\) multiply by 5: \(5(F - 32) = 9C\)
divide by 9: \(30 = C\) divide by 9: \(\dfrac{5(F - 32)}{9} = C\)

The steps are identical. On the right, instead of getting one number, we get a new formula:

\[C = \frac{5(F - 32)}{9}.\]

Now 59 °F is \(\tfrac{5 \times 27}{9} = 15\) °C, in one line.

Undoing in reverse order

A formula is like a machine: it takes the starting quantity through a chain of operations. To find the starting quantity again, you undo the operations in reverse order, like taking off your shoes before your socks.

CF CF →→→ ←←← × 9/5+ 32 × 5/9− 32 forwards: work out Fbackwards: undo in reverse order

Going forwards, \(C\) is multiplied by \(\tfrac95\) first and 32 is added last. So going backwards, you undo the last thing first: subtract 32, then multiply by \(\tfrac59\) (the opposite of multiplying by \(\tfrac95\)).

To see the order of operations in a formula, ask "if I knew the subject's partner, what would I work out first?" In \(v = u + at\) (speed \(v\) after accelerating at rate \(a\) for time \(t\), starting at speed \(u\)), to find \(v\) you would first multiply \(a\) by \(t\), then add \(u\). So to make \(t\) the subject, first subtract \(u\), then divide by \(a\):

\[v - u = at \quad\Rightarrow\quad t = \frac{v - u}{a}.\]

Watch the brackets

When you divide a whole side by something, the whole side must be divided. So the answer above is \(t = \dfrac{v - u}{a}\), with the fraction bar under all of \(v - u\). Writing \(t = v - \dfrac{u}{a}\) would divide only \(u\), and that is a different formula. Check with numbers: if \(u = 5\), \(v = 25\) and \(a = 4\), the right formula gives \(t = \tfrac{20}{4} = 5\), and indeed \(5 + 4 \times 5 = 25\). ✓ The wrong one gives \(25 - 1.25 = 23.75\).

Squares and square roots

In \(A = \pi r^2\) (the area of a circle) the last thing done to \(r\) is squaring it, after which it is multiplied by \(\pi\). Undo in reverse: divide by \(\pi\), then take the square root.

\[\frac{A}{\pi} = r^2 \quad\Rightarrow\quad r = \sqrt{\frac{A}{\pi}}.\]

Strictly, \(r^2 = \tfrac{A}{\pi}\) has two solutions, a positive and a negative square root. But \(r\) is a length, so only the positive one makes sense. Always ask what the letters mean: it tells you which answers to keep.

When the subject appears twice

In \(P = 2l + 2w\) (the perimeter of a rectangle) the subject \(w\) appears once, so it is straightforward: \(P - 2l = 2w\), so \(w = \dfrac{P - 2l}{2}\).

But sometimes the letter you want appears in two places, as in \(ax + b = cx + d\) when you want \(x\). Then you gather all the \(x\) terms on one side, exactly as in the last lesson, and use the distributive law backwards to pull \(x\) out as a common factor:

\[ \begin{aligned} ax + b &= cx + d\\ ax - cx &= d - b\\ x(a - c) &= d - b\\ x &= \frac{d - b}{a - c}. \end{aligned} \]

Things you must not divide by

The last line divided by \(a - c\). That is only allowed if \(a - c\) is not zero, because dividing by zero has no meaning. So the full answer is "\(x = \tfrac{d - b}{a - c}\), as long as \(a \ne c\)". The same caution applies to \(t = \tfrac{v - u}{a}\): if \(a = 0\) the speed never changes, and there is no time to find. Good mathematicians note these restrictions, and they usually match something real about the situation.

Check by putting numbers in

A rearranged formula can always be tested. Pick easy numbers that fit the original formula, and see whether the new one gives them back. For \(C = \tfrac{5(F - 32)}{9}\): water boils at 100 °C, which is 212 °F, and \(\tfrac{5 \times 180}{9} = 100\). ✓

Worked examples

1. The size of Mars. Mars's equator is about 21,340 km around. Make \(r\) the subject of \(C = 2\pi r\) and find the radius of Mars at its equator.

  • The formula multiplies \(r\) by \(2\pi\). Undo it by dividing both sides by \(2\pi\): \(r = \dfrac{C}{2\pi}\).
  • \(r = \dfrac{21340}{2\pi} \approx 3396\) km. That is a bit over half of Earth's radius (about 6378 km at the equator).

2. The forecast. Make \(C\) the subject of \(F = \tfrac95 C + 32\), then convert 86 °F and −40 °F.

  • From the table above, \(C = \dfrac{5(F - 32)}{9}\).
  • 86 °F: \(C = \dfrac{5 \times 54}{9} = \dfrac{270}{9} = 30\) °C. A hot summer day.
  • −40 °F: \(C = \dfrac{5 \times (-72)}{9} = -40\) °C. The one temperature where both scales agree.

3. A speed formula. Make \(t\) the subject of \(v = u + at\). A cyclist speeds up from 5 m/s to 25 m/s with acceleration 4 m/s every second. How long does it take?

  • Subtract \(u\): \(v - u = at\). Divide by \(a\): \(t = \dfrac{v - u}{a}\).
  • \(t = \dfrac{25 - 5}{4} = 5\) seconds. Check: \(5 + 4 \times 5 = 25\). ✓

4. Two steps with a fraction. Make \(h\) the subject of \(A = \tfrac12 bh\) (the area of a triangle).

  • Multiply both sides by 2: \(2A = bh\).
  • Divide both sides by \(b\): \(h = \dfrac{2A}{b}\).
  • Check: a triangle with base 6 and height 4 has area 12, and \(\tfrac{2 \times 12}{6} = 4\). ✓

5. The subject twice. Make \(x\) the subject of \(ax = bx + c\).

  • Subtract \(bx\) from both sides: \(ax - bx = c\).
  • Factor out \(x\): \(x(a - b) = c\).
  • Divide by \(a - b\) (allowed if \(a \ne b\)): \(x = \dfrac{c}{a - b}\).
  • Check with \(a = 5\), \(b = 2\), \(c = 12\): \(x = \tfrac{12}{3} = 4\), and indeed \(5 \times 4 = 2 \times 4 + 12\). ✓

Try it

  1. Make \(s\) the subject of \(P = 4s\) (the perimeter of a square).
  2. Make \(x\) the subject of \(y = 3x - 7\).
  3. Make \(b\) the subject of \(A = \tfrac12 bh\).
  4. Make \(l\) the subject of \(P = 2l + 2w\), then find \(l\) when \(P = 30\) and \(w = 4\).
  5. Use the formulas in this lesson to convert 20 °C into Fahrenheit and 50 °F into Celsius.
  6. The distance travelled while speeding up steadily is \(s = \tfrac12(u + v)t\). Make \(v\) the subject.
  7. Make \(r\) the subject of \(V = \pi r^2 h\) (the volume of a cylinder).
  8. Make \(x\) the subject of \(5x + a = 2x + b\). What is \(x\) when \(a = 1\) and \(b = 13\)?
Answers
  1. Divide by 4: \(s = \dfrac{P}{4}\).
  2. Add 7: \(y + 7 = 3x\). Divide by 3: \(x = \dfrac{y + 7}{3}\). Check with \(x = 2\): \(y = -1\), and \(\tfrac{-1 + 7}{3} = 2\). ✓
  3. Multiply by 2: \(2A = bh\). Divide by \(h\): \(b = \dfrac{2A}{h}\).
  4. Subtract \(2w\): \(P - 2w = 2l\). Divide by 2: \(l = \dfrac{P - 2w}{2}\) (which is also \(\tfrac{P}{2} - w\)). With \(P = 30\), \(w = 4\): \(l = \tfrac{22}{2} = 11\). Check: \(2 \times 11 + 2 \times 4 = 30\). ✓
  5. \(F = \tfrac95 \times 20 + 32 = 36 + 32 = 68\) °F. \(C = \tfrac{5(50 - 32)}{9} = \tfrac{90}{9} = 10\) °C.
  6. Multiply by 2: \(2s = (u + v)t\). Divide by \(t\): \(\tfrac{2s}{t} = u + v\). Subtract \(u\): \(v = \dfrac{2s}{t} - u\). Check with \(u = 2\), \(v = 6\), \(t = 3\): \(s = \tfrac12 \times 8 \times 3 = 12\), and \(\tfrac{24}{3} - 2 = 6\). ✓
  7. Divide by \(\pi h\): \(r^2 = \dfrac{V}{\pi h}\). Take the positive square root (a radius is a length): \(r = \sqrt{\dfrac{V}{\pi h}}\).
  8. Subtract \(2x\) and \(a\): \(3x = b - a\), so \(x = \dfrac{b - a}{3}\). With \(a = 1\), \(b = 13\): \(x = 4\). Check: \(20 + 1 = 21\) and \(8 + 13 = 21\). ✓

Common mistake

Dividing only part of a side. From \(v - u = at\), dividing by \(a\) gives \(t = \tfrac{v - u}{a}\), with the whole of \(v - u\) on top. Writing \(v - \tfrac{u}{a}\) is a different formula. The fix: put brackets round the whole side before you divide, \(t = (v - u) \div a\), and then write it as a fraction.

Undoing in the wrong order. For \(F = \tfrac95 C + 32\), dividing by \(\tfrac95\) first and then subtracting 32 gives \(C = \tfrac59 F - 32\), which turns 212 °F into about 85.8 °C instead of 100 °C. The 32 was added last, so it must be taken away first. Testing with a pair of numbers you know, like boiling water, catches this at once.

Where it leads