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

Starter quiz

In which of these formulae is $a$ the subject?

$a + 1 = bc$

$\frac{b+c}{3}=a$ ✓

$a^2=c^2-b^2$

$bc + 5 = \frac{a}{4}$

$a=\frac{2b + 5}{c(2-b)} + d$ ✓
Make $x$ the subject of $3x = y + z$ .

$x=3(y+z)$

$x=y+z-2x$

$x = \frac{y}{3} + z$

$x = \frac{y+z}{3}$ ✓
Make $a$ the subject of $4-ab=c$ .

$a={4\over bc}$

$a={4-c\over b}$ ✓

$a={4+c\over b}$

$a={c-4\over b}$

$-a={4-c\over b}$
The solution to the equation ${10\over x} = 2$ is when $x=$ ______.

'5' ✓
What is the solution to the equation ${5\over x}=7$ ?

$x={5\over 7}$ ✓

$x={7\over 5}$

$x = 2$

$x = 35$
You are given the formula $w=2xy+3x^2$ . When $x=3$ and $y = 4$ , $w=$ ______.

'51' ✓

Exit quiz

Make $x$ the subject of $y = 2\pi x$ .

$x = \frac{y}{2}$

$x = \frac{y}{2\pi}$ ✓

$x = \frac{2y}{\pi}$

$x = \frac{\pi y}{2}$

$x = y - 2\pi$
Make $h$ the subject of $sin(\theta )={o\over h}$ .

$h=o \times sin(\theta )$

$h = \frac{sin(\theta )}{o}$

$h = \frac{o}{sin(\theta )}$ ✓

$\frac{1}{h} = \frac{sin(\theta )}{o}$
Make $a$ the subject of $b=\frac{5}{a+c}$ .

$a = {5\over bc}$

$a = {5\over b}+c$

$a = {5\over b+c}$

$a = {5-c\over b}$

$a = {5-bc\over b}$ ✓
Make $r$ the subject of $p = \sqrt {q\over r}+1$ .

$r = {q\over (p-1)^2}$ ✓

$r = {q\over p^2-1}$

$r = {p^2\over q^2}-1$

$r = {(p-1)^2\over q}$

$r = {q\over \sqrt {p+1}}$
A formula for calculating the perimeter of a rectangle is shown. When $P= 16$ and $b = 5$ , the value of $a$ is ______.

'3' ✓
$The formula for the area of a sector is shown. When the area, <Math>A</Math>, is 110 cm² and the radius, <Math>r</Math>, is 10 cm, the value of <Math>\theta</Math> is <span class="blank">______</span> to the nearest degree.$

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.

'126' ✓

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Lesson Details

Key learning points

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

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.

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