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.
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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.
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.
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.
To help you plan your year 9 maths lesson on: Changing the subject with more complex formula, download all teaching resources for free and adapt to suit your pupils' needs...
To help you plan your year 9 maths lesson on: Changing the subject with more complex formula, download all teaching resources for free and adapt to suit your pupils' needs.
The starter quiz will activate and check your pupils' prior knowledge, with versions available both with and without answers in PDF format.
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Explore more key stage 3 maths lessons from the Expressions and formulae unit, dive into the full secondary maths curriculum, or learn more about lesson planning.
Licence
Lesson 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
$${{d+6}\over8}=e$$ -
$$d=8e-6$$
$$8d+6=e$$ -
$${{e-6}\over8}=d$$
$${{d-6}\over8}=e$$ -
$$d=8e+6$$
$$8(d+6)=e$$ -
$${e\over8}-6=d$$
$${d\over8}-6=e$$ -
$$d=8(e+6)$$
$$2x=\sqrt{y}$$ -
$$x={\sqrt{y}\over2}$$
$$2\sqrt{x}=y$$ -
$$x=({y\over2})^2$$
$$\sqrt{2x}=y$$ -
$$x={y^2\over2}$$
$$2x^2=y$$ -
$$x=\sqrt{y\over2}$$
$$(2x)^2=y$$ -
$$x={{\sqrt{y}}\over2}$$