Changing the Subject of a Formula

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Changing the Subject of a Formula

The best way to learn and improve is to get right into it so let’s start with a simple exercise.

Getting Started on Changing the Subject of a Formula

Exam Tip
The inverse of an operation will undo the operation.
  • The inverse of multiply is divide.
  • The inverse of divide is multiply.
  • The inverse of add is subtract.
  • The inverse of subtract is add.
  • The following examples contain everything you need to know about changing the subject of a formula.

    This is very similar to solving linear equations so you may want to revise the two topics together.

    Example 1
    Make 𝑝 the subject of π‘₯ = 4𝑝 + 𝑦.

    Making 𝑝 the subject means we want a formula in the form 𝑝 = _. We do this by using inverse operations to manipulate the formula until we have 𝑝 on its own.

    First, we want to get the term including 𝑝 on its own. We can do this by subtracting 𝑦 from both sides.

    π‘₯ = 4𝑝 + 𝑦
    – 𝑦– 𝑦
    π‘₯ – 𝑦 = 4𝑝
    Γ· 4Γ· 4
    \frac{x-y}{4} \ = 𝑝

    It doesn’t matter that we have 𝑝 on the right-hand side rather than the left-hand side. Both sides of the formula are equal, so 𝑝 is the subject of this formula. If you prefer, you can swap the sides and write the formula as 𝑝 = \frac{x - y}{4} but there is no need to.

    Example 2
    Make π‘˜ the subject of π‘˜ 3 – 𝑝 = π‘š

    This example is very similar to example 3. The key difference is the order in which we perform our manipulation. In this case, π‘˜ is divided by 3 and then 𝑝 is subtracted. To change the subject, we need to undo each of these in reverse order.

    \frac{k}{3} \ – 𝑝 = π‘š
    + 𝑝+ 𝑝
    \frac{k}{3} \ = π‘š + 𝑝
    Γ— 3Γ— 3
    π‘˜ = 3(π‘š + 𝑝)
    Example 3
    Make π‘˜ the subject of π‘˜ – 𝑝 3 = π‘š

    This example is very similar to example 2. The key difference is the order in which we perform our manipulation. In this case, we subtract 𝑝 and then divide by 3. To change the subject we need to undo each of these in reverse order.

    Exam Tip
  • The inverse of squaring is square rooting
  • The inverse of square rooting is squaring
  • \frac{k-p}{3} \ = π‘š
    Γ— 3Γ— 3
    π‘˜ – 𝑝 = 3π‘š
    + 𝑝+ 𝑝
    π‘˜ = 3π‘š + 𝑝
    Example 4
    Make π‘₯ the subject of 𝑏 = \sqrt{x^2 - 3a}

    This time, we need to start by eliminating the square root.

    𝑏 = \sqrt{x^2 - 3a} \
    _2_2
    𝑏2 = π‘₯2 – 3π‘Ž
    + 3π‘Ž+ 3π‘Ž
    𝑏2 + 3π‘Ž = π‘₯2
    \sqrt{\textunderscore} \sqrt{\textunderscore}
    \sqrt{b^2 + 3a} \ = π‘₯

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