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Changing the Subject of a Formula

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Making a letter the subject means undoing what is done to it, both sides at once

The subject of a formula

In the formula y = mx + c, y is the subject: it stands alone on one side, and the formula tells you how to work it out from x, m and c. Changing the subject means rearranging the formula so that a different letter stands alone.

A formula is an equation, so the rule is the one you use for solving equations: whatever you do to one side, do to the other. The difference is that the answer is an expression in the other letters, not a number.

What happens to x

Follow x through the formula y = mx + c. First x is multiplied by m, giving mx. Then c is added, giving y. To get x back from y, undo these steps in the reverse order: first undo the adding, then undo the multiplying.

x× mmx+ cy

Going forward, x is multiplied by m, and then c is added.

y− cy − c÷ mx

Going back, the last step is undone first: subtract c, and then divide by m.

Rearranging, one step at a time

Start with y = mx + c. Subtract c from both sides: y − c = mx.

Now divide both sides by m. The whole of the left side, y − c, is divided by m, not just the y: . The bracket shows that the subtraction happens before the division.

It is usual to write the subject first, so turn the formula round: . It is the same formula as y = mx + c, rearranged so that x stands alone.

Using the new formula

Take y = 3x + 5. Here m = 3 and c = 5, so . When y = 17, . Check in the original formula: 3 × 4 + 5 = 17.

The letter you want decides the steps

A formula can be rearranged for any of its letters. To make m the subject of y = mx + c, start the same way: subtract c from both sides, y − c = mx. This time m is multiplied by x, so divide both sides by x: .

Another example: in v = u + at, make t the subject. Subtract u from both sides: v − u = at. Divide both sides by a: .

Undo in the reverse order

The area of a circle is . The radius is squared first, then multiplied by . To make r the subject, undo those steps last one first: divide both sides by , so , then take the square root, so . Taking the square root first undoes the steps in the wrong order and gives a different number.

r = 3A = 28.3A = πr²r = 3 → ² → 9 → × π → A = 28.3A = 28.3 → √ → 5.32 → ÷ π → 1.69 ≠ r ✗÷ π, √√, ÷ π

Undo the steps in the order that brings r back

Choose the order of the two undo steps. With first and then the square root, r comes back, and dragging the radius shows it comes back for every circle.

The usual mistakes

Dividing only part of a side. From y − c = mx, dividing by m gives , not . Every term on the side is divided.

Dividing by m first and forgetting the c. Dividing both sides of y = mx + c by m divides every term: , not x + c. Subtracting c first, while it is still a separate term, avoids the fraction .

Equations and inequalities · The Missing Parallel Side of a Trapezium-Shaped Wall

QuestionThe end wall of a shed is a trapezium. The area of a trapezium is given by the formula A = 12(a + b)h, where a and b are the lengths of the two parallel sides and h is the distance between them. (a) Make b the subject of the formula. (b) The wall has an area of 42 m2. One of its parallel sides is 5 m long and the distance between the parallel sides is 6 m. Find the length of the other parallel side.

  1. 1.Start from A = 12(a + b)h. Multiply both sides by 2 to remove the fraction: 2A = (a + b)h.

    abhAA=1/2 (a + b)h2A=(a + b)hmultiply both sides by 2
    Multiply both sides of A = 12(a + b)h by 2: 2A = (a + b)h.
  2. 2.Divide both sides by h, so that the bracket stands alone: 2Ah = a + b.

    abhAA=1/2 (a + b)h2A=(a + b)hmultiply both sides by 22A/h=a + bdivide both sides by h
    Divide both sides by h, so that the bracket stands alone: 2Ah = a + b.
  3. 3.Subtract a from both sides: 2Ah − a = b. (a) b = 2Ah − a.

    abhAA=1/2 (a + b)h2A=(a + b)hmultiply both sides by 22A/h=a + bdivide both sides by h2A/h − a=bsubtract a from both sides
    (a) Subtract a from both sides: b = 2Ah − a.
  4. 4.Substitute A = 42, h = 6 and a = 5: b = 2 × 426 − 5 = 846 − 5 = 14 − 5 = 9.

    a = 5b = ?h = 6A = 42A=1/2 (a + b)h2A=(a + b)hmultiply both sides by 22A/h=a + bdivide both sides by h2A/h − a=bsubtract a from both sidesb = 84/6 − 5 = 14 − 5 = 9
    Substitute A = 42, h = 6 and a = 5: b = 846 − 5 = 9.
  5. 5.(b) The other parallel side is 9 m long. Check with the original formula: 12 × (5 + 9) × 6 = 12 × 14 × 6 = 42 m2.

    a = 5b = 9h = 6A = 42A=1/2 (a + b)h2A=(a + b)hmultiply both sides by 22A/h=a + bdivide both sides by h2A/h − a=bsubtract a from both sidesb = 84/6 − 5 = 14 − 5 = 9check: 1/2 × (5 + 9) × 6 = 42
    (b) The other parallel side is 9 m long.

Answer: (a) b = 2Ah − a; (b) 9 m

Watch out for

  • Subtracting a first and writing A − a = 12bh. The side a is inside the bracket, so it is multiplied by 12h as well. The bracket must stand alone before a is subtracted.
  • Working out 2 × 426 − 5. Only 2A is divided by h, and a is subtracted afterwards: 846 − 5 = 9.

Your turn

1.Make x the subject of y = 2x + 3 

y = 2x + 3which move frees x?
Practice Changing the Subject of a Formula in the app