Determine the missing quantities in your study schedule with Beyond Revision and tuition-style posts like this one on 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 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 π the subject of π₯ = 4π + π¦.

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

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

π₯ = 4π + π¦ | ||

– π¦ | – π¦ | |

π₯ – π¦ = 4π | ||

Γ· 4 | Γ· 4 | |

= π |

It doesnβt matter that we have π on the right-hand side rather than the left-hand side. Both sides of the formula are equal, so π is the subject of this formula. If you prefer, you can swap the sides and write the formula as π = but there is no need to.

**Example 2**

Make π the subject of $\frac{\mathrm{\pi \x9d\x91\x98}}{3}$ – π = π

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 π is subtracted. To change the subject, we need to undo each of these in **reverse order**.

– π = π | ||

+ π | + π | |

= π + π | ||

Γ 3 | Γ 3 | |

π = 3(π + π) |

**Example 3**

Make π the subject of $\frac{\mathrm{\pi \x9d\x91\x98\; \u2013\; \pi \x9d\x91\x9d}}{3}$ = π

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 *π* and then divide by 3. To change the subject we need to undo each of these in reverse order.

**Exam Tip**

**squaring**is

**square rooting**

**square rooting**is

**squaring**

= π | ||

Γ 3 | Γ 3 | |

π – π = 3π | ||

+ π | + π | |

π = 3π + π |

**Example 4**

Make π₯ the subject of π =

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

π = | ||

_^{2} | _^{2} | |

π^{2} = π₯^{2} – 3π | ||

+ 3π | + 3π | |

π^{2} + 3π = π₯^{2} | ||

= π₯ |

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