# Changing the subject with more complex formula

I can apply an understanding of inverse operations to a complex formula in order to make a specific variable the subject.

# Changing the subject with more complex formula

I can apply an understanding of inverse operations to a complex formula in order to make a specific variable the subject.

## Lesson details

### Key learning points

- With more complex formula, it is important to apply the inverse operations in the right order.
- The subject can be thought of as the unknown you are trying to find the value of.
- Instead of finding a value though, you will find an expression that the unknown is equal to.

### Common misconception

The subject of a formula is just the first term in the formula.

Draw attention to the variety of equations and formulae used in the lesson.

### Keywords

Subject of an equation/formula - The subject of an equation/a formula is a variable that is expressed in terms of other variables. It should have an exponent of 1 and a coefficient of 1.

### Licence

This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

## Video

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## Starter quiz

### 6 Questions

$$5d+2$$ -

$$d \text{ }$$ multiplied by $$5$$ then add $$2$$

$${d\over5}+2$$ -

$$d\text{ }$$ divided by $$5$$ then add $$2$$

$${{d+5}\over2}$$ -

$$d\text{ }$$ add $$5$$ then divided by $$2$$

$$2d+5$$ -

$$d\text{ }$$ multiplied by $$2$$ then add $$5$$

$$5(d+2)$$ -

$$d\text{ }$$ add $$2$$ then multiplied by $$5$$

$${1\over5}(d+2)$$ -

$$d\text{ }$$ add $$2$$ then divided by $$5$$

## Exit quiz

### 6 Questions

$${{d+6}\over8}=e$$ -

$$d=8e-6$$

$$8d+6=e$$ -

$${{e-6}\over8}=d$$

$${{d-6}\over8}=e$$ -

$$d=8e+6$$

$$8(d+6)=e$$ -

$${e\over8}-6=d$$

$${d\over8}-6=e$$ -

$$d=8(e+6)$$

$$2x=\sqrt{y}$$ -

$$x={\sqrt{y}\over2}$$

$$2\sqrt{x}=y$$ -

$$x=({y\over2})^2$$

$$\sqrt{2x}=y$$ -

$$x={y^2\over2}$$

$$2x^2=y$$ -

$$x=\sqrt{y\over2}$$

$$(2x)^2=y$$ -

$$x={{\sqrt{y}}\over2}$$