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Year 11
Foundation

Checking and securing understanding of changing the subject with simple algebraic fractions

I can change the subject when a simple algebraic fraction is present in an equation or formula.

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New
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Year 11
Foundation

Checking and securing understanding of changing the subject with simple algebraic fractions

I can change the subject when a simple algebraic fraction is present in an equation or formula.

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Lesson details

Key learning points

  1. It is important to apply the inverse operations in the right order.
  2. The subject can be thought of as the unknown you are trying to find the value of.
  3. 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

When rearranging multiplicative relationships, the values just 'swap'.

Pupils should understand how to manipulate multiplicative relationships properly instead of relying on a 'trick'. This will be useful in other subject areas as well.

LC2 provides an opportunity to revisit the area of a sector. Pupils could try to write their own formula first. Discuss whether changing the subject or substituting first is the best way to solve questions where they need to find the angle. Substituting first is often easier for foundation pupils.
Teacher tip

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).

Lesson video

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Starter quiz

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6 Questions

Q1.
In which of these formulae is $$a$$ the subject?
$$a + 1 = bc$$
Correct answer: $$\frac{b+c}{3}=a$$
$$a^2=c^2-b^2$$
$$bc + 5 = \frac{a}{4}$$
Correct answer: $$ a=\frac{2b + 5}{c(2-b)} + d$$
Q2.
Make $$x$$ the subject of $$3x = y + z$$.
$$x=3(y+z)$$
$$x=y+z-2x$$
$$x = \frac{y}{3} + z$$
Correct answer: $$x = \frac{y+z}{3}$$
Q3.
Make $$a$$ the subject of $$4-ab=c$$.
$$a={4\over bc}$$
Correct answer: $$a={4-c\over b}$$
$$a={4+c\over b}$$
$$a={c-4\over b}$$
$$-a={4-c\over b}$$
Q4.
The solution to the equation $${10\over x} = 2$$ is when $$x=$$ .
Correct Answer: 5
Q5.
What is the solution to the equation $${5\over x}=7$$?
Correct answer: $$x={5\over 7}$$
$$x={7\over 5}$$
$$x = 2$$
$$x = 35$$
Q6.
You are given the formula $$w=2xy+3x^2$$. When $$x=3$$ and $$y = 4$$, $$w=$$ .
Correct Answer: 51

Exit quiz

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6 Questions

Q1.
Make $$x$$ the subject of $$y = 2\pi x$$.
$$x = \frac{y}{2}$$
Correct answer: $$x = \frac{y}{2\pi}$$
$$x = \frac{2y}{\pi}$$
$$x = \frac{\pi y}{2}$$
$$x = y - 2\pi $$
Q2.
Make $$h$$ the subject of $$sin(\theta )={o\over h}$$.
$$h=o \times sin(\theta )$$
$$h = \frac{sin(\theta )}{o}$$
Correct answer: $$h = \frac{o}{sin(\theta )}$$
$$\frac{1}{h} = \frac{sin(\theta )}{o}$$
Q3.
Make $$a$$ the subject of $$b=\frac{5}{a+c}$$.
$$a = {5\over bc}$$
$$ a = {5\over b}+c$$
$$a = {5\over b+c}$$
$$a = {5-c\over b}$$
Correct answer: $$a = {5-bc\over b}$$
Q4.
Make $$r$$ the subject of $$ p = \sqrt {q\over r}+1$$.
Correct answer: $$r = {q\over (p-1)^2}$$
$$r = {q\over p^2-1}$$
$$r = {p^2\over q^2}-1$$
$$r = {(p-1)^2\over q}$$
$$r = {q\over \sqrt {p+1}}$$
Q5.
A formula for calculating the perimeter of a rectangle is shown. When $$P= 16$$ and $$b = 5$$, the value of $$a$$ is .
An image in a quiz
Correct Answer: 3, a=3
Q6.
The formula for the area of a sector is shown. When the area, $$A$$, is 110 cm² and the radius, $$r$$, is 10 cm, the value of $$\theta$$ is to the nearest degree.
An image in a quiz
Correct Answer: 126, 126 degrees