Making a Twice-Appearing Letter the Subject

Gather, factor, then share.

When the letter appears twice

Make x the subject of ax = bx + c. There is an x term on each side, so dividing by a does not work: it gives x = (bx + c)/a, and x is still on the right. A formula for x must have x on one side only.

The method has three steps: gather every x term on one side, factor x out, then divide.

Gather the x terms

This is the same move as solving an equation with x on both sides. Subtract bx from both sides: ax − bx = c. Now both x terms are on the left, and c is on its own on the right.

axbx + cboth sides − bx

Take bx off both pans. The left keeps ax − bx, and the right keeps c.

Factor x out

ax − bx has two terms, and x is a factor of both. Take x out as a common factor: ax − bx = x(a − b). The two x terms have become one x, multiplied by the bracket (a − b).

So the formula is now x(a − b) = c.

x(a − b)bxa − bbxax − bx = x(a − b)

ax is a rectangle x tall and a long. Take away the strip bx, and what is left is x tall and a − b long: x(a − b).

Divide by the whole bracket

x is multiplied by (a − b), so divide both sides by (a − b): x = c/(a − b). Now x appears once, on its own, and the formula is finished.

Divide by the whole bracket, not by a alone or by b alone. And the formula works only when a − b is not 0: if a = b, there is nothing to divide by.

Check with numbers. If a = 5, b = 2 and c = 12, the formula gives x = 12/(5 − 2) = 12/3 = 4. In the original, 5 × 4 = 20 and 2 × 4 + 12 = 20.

A harder example

Make x the subject of px + q = rx + s. Gather the x terms on the left and the other terms on the right: subtract rx from both sides, then subtract q from both sides. This gives px − rx = s − q.

Factor x out of the left side: x(p − r) = s − q. Divide both sides by (p − r): x = (s − q)/(p − r).

The usual mistakes

Stopping while x is still on both sides. x = (bx + c)/a is true, but it is not a formula for x, because you need x to use it.

Dividing by only part of the bracket. From x(a − b) = c, the answer is x = c/(a − b). Writing x = c/a − b divides c by a and then subtracts b, which is a different formula.

Worked example: Two Resistors in Parallel and the Second Resistor a Circuit Needs

Question When two resistors of R1 ohms and R2 ohms are connected in parallel, their combined resistance R ohms is given by R = R1 R2R1 + R2. (a) Make R2 the subject of the formula. (b) A technician has a 60 ohm resistor and needs a combined resistance of 24 ohms. Find the resistance of the second resistor she must connect in parallel with it. Then use your formula to explain why no resistor connected in parallel with the 60 ohm one gives a combined resistance of 75 ohms.

  1. 1.Multiply both sides by R1 + R2 to clear the fraction: R(R1 + R2) = R1 R2. Expand the bracket: R R1 + R R2 = R1 R2.

    R=R1 × R2/(R1 + R2)R(R1 + R2)=R1 × R2multiply both sides by R1 + R2R × R1 + R × R2=R1 × R2expand the bracket
    R=R1 × R2/(R1 + R2)multiply both sides by R1 + R2R(R1 + R2)=R1 × R2expand the bracketR × R1 + R × R2=R1 × R2
    Multiply both sides by R1 + R2 to clear the fraction, then expand: R R1 + R R2 = R1 R2.

  2. 2.R2 is in two terms. Gather them on one side by subtracting R R2 from both sides: R R1 = R1 R2 − R R2.

    R=R1 × R2/(R1 + R2)R(R1 + R2)=R1 × R2multiply both sides by R1 + R2R × R1 + R × R2=R1 × R2expand the bracketR × R1=R1 × R2 − R × R2subtract R × R2 from both sides
    R=R1 × R2/(R1 + R2)multiply both sides by R1 + R2R(R1 + R2)=R1 × R2expand the bracketR × R1 + R × R2=R1 × R2subtract R × R2 from both sidesR × R1=R1 × R2 − R × R2
    Gather the two terms in R2 on one side: subtract R R2 from both sides.

  3. 3.Factor out R2: R R1 = R2(R1 − R). Divide both sides by R1 − R. (a) R2 = R R1R1 − R.

    R=R1 × R2/(R1 + R2)R(R1 + R2)=R1 × R2multiply both sides by R1 + R2R × R1 + R × R2=R1 × R2expand the bracketR × R1=R1 × R2 − R × R2subtract R × R2 from both sidesR × R1=R2(R1 − R)factor out R2R × R1/(R1 − R)=R2divide both sides by R1 − R
    R=R1 × R2/(R1 + R2)multiply both sides by R1 + R2R(R1 + R2)=R1 × R2expand the bracketR × R1 + R × R2=R1 × R2subtract R × R2 from both sidesR × R1=R1 × R2 − R × R2factor out R2R × R1=R2(R1 − R)divide both sides by R1 − RR × R1/(R1 − R)=R2
    (a) Factor out R2, then divide both sides by R1 − R: R2 = R R1R1 − R.

  4. 4.Substitute R1 = 60 and R = 24: R2 = 24 × 6060 − 24 = 144036 = 40. Check in the original formula: 60 × 4060 + 40 = 2400100 = 24 ohms.

    60 ohmsR2 = 40 ohmscombined: R = 24 ohmsR2 = 24 × 60/(60 − 24) = 1440/36 = 40check: 60 × 40/(60 + 40) = 2400/100 = 24
    60 ohmsR2 = 40 ohmscombined: R = 24 ohmsR2 = 24 × 60/(60 − 24) = 1440/36 = 40check: 60 × 40/(60 + 40) = 2400/100 = 24
    With R1 = 60 and R = 24: R2 = 144036 = 40 ohms. Check: 60 × 40100 = 24.

  5. 5.Substitute R = 75: R2 = 75 × 6060 − 75 = 4500−15 = −300. A resistance cannot be negative, so this value is rejected.

    60 ohmsR2 = ?combined: R = 75 ohmsR2 = 75 × 60/(60 − 75)= 4500/(−15) = −300 ohmsa resistance cannot be negative: rejected
    60 ohmsR2 = ?combined: R = 75 ohmsR2 = 75 × 60/(60 − 75)= 4500/(−15) = −300 ohmsa resistance cannot be negative: rejected
    With R = 75: R2 = 4500−15 = −300. A resistance cannot be negative, so this is rejected.

  6. 6.(b) The second resistor is 40 ohms. No resistor gives 75 ohms: R2 is positive only when R1 − R is positive, so the combined resistance is always less than 60 ohms.

    60 ohmsR2 = 40 ohmsR2 is positive only when R1 − R is positiveso R is always less than 60 ohmsR2 = 40 ohms gives R = 24 ohms
    60 ohmsR2 = 40 ohmsR2 is positive only when R1 − R is positiveso R is always less than 60 ohmsR2 = 40 ohms gives R = 24 ohms
    (b) The second resistor is 40 ohms. R2 is positive only when R is less than R1, so no resistor gives 75 ohms.

Answer: (a) R2 = R R1R1 − R; (b) 40 ohms; 75 ohms would need R2 = −300 ohms, which is impossible

Common mistakes

  • Stopping at R2 = R(R1 + R2)R1. That is not a formula for R2, because R2 is still on the right. The terms in R2 must be gathered on one side and R2 factored out.
  • Taking −300 ohms as the resistor to use, or as a slip in the arithmetic. The formula gives a negative number because the request is impossible: two resistors in parallel always have less resistance than either one alone.

More equations and inequalities problems, worked step by step →

Practice Making a Twice-Appearing Letter the Subject in the app