Indicies - Explanation, Examples and Laws of Indicies

                                            Change of Subject of a Relation

Introduction

In algebra, you often meet formulas where one quantity depends on others — like

$$A = πr^2$$$$v=u+at$$

But sometimes, you’re asked to make another letter the subject. For example, instead of $A = πr^2$, you might need to make $\mathrm{r\; the\; subject.}$

This process is called “Change of Subject of a Relation.”
It simply means rearranging a formula so that one specific variable stands alone on one side of the equation- usually the left-hand side.

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What It Means

If we have an equation: y = 3x+2 and we want to make x the subject, it means we must rewrite it as:

$$x=\text{something in terms of }y$$

It’s like solving for $x$ — isolating it by performing the same operations on both sides.

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Basic Steps for Changing the Subject

1. Identify the variable you want to make the subject.

2. Undo all operations attached to it step by step (in reverse order of the BODMAS rule).

3. Keep the equation balanced by performing the same operation on both sides.

4. Simplify where possible.

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Step-by-Step Examples

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Example 1: y=2x+5

Make x the subject.

Solution: y=2x+5

Subtract 5 from both sides: 

y−5= 2x-5

=> y-5=2x

Divide both sides by 2:

$$x=\frac{y-5}{\mathrm{2}}$$

Answer: $x = \frac{y - 5}{2}$

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Example 2:

$$A = πr^2$$

Make r the subject.

Solution:

$$A = πr^2$$

Divide both sides by π:

$$\frac{A}{π} = r^2$$

Take square root of both sides:

$$r = \sqrt{\frac{A}{π}}$$

Answer: $r = \sqrt{\frac{A}{π}}$

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Example 3:

$$v = u + at$$

Make t the subject.

Solution:

We'll subtract  'u' from both sides of the equation

$$v - u = at$$

Divide both sides by $a$:

$$t = \frac{v - u}{a}$$

Answer: $t = \frac{v - u}{a}$

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Example 4:

$$P = \frac{2L + 2B}{1}$$

Make L the subject.

Solution:

$$P = 2L + 2B$$

Subtract $2B$ from both sides:

$$P - 2B = 2L$$

Divide both sides by 2:

$$L=\frac{P-2B}{\mathrm{2}}$$

Answer: $L = \frac{P - 2B}{2}$

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Example 5:

$$s = \frac{u + v}{2}t$$

Make v the subject.

Solution:

$$\[ s=\frac{u+v}{2} \]\mathrm{ x }\frac{t<br>}{1}$$

$$\frac{\mathrm{<br>}}{}$$$$\mathrm{=>\; s}=\frac{\mathrm{t(u}+\mathrm{v)}}{2}$$

Multiply both sides by 2:

$$2s=t(u+v)$$

Divide both sides by $t$:

$$\frac{2s}{t}=u+v$$

Subtract $u$ from both sides:

$$v = \frac{2s}{t} - u$$

Answer: $v = \frac{2s}{t} - u$

Example 6:

$$y=\frac{3x-4}{\mathrm{5}}$$

Make x the subject.

Solution:
Multiply both sides by 5:

$$5y=3x-4\mathrm{<br>}$$

Add 4 to both sides:

(NB: This is to eliminate or nullify the number to make 'x' stand alone since we're trying to make it the subject)

$$5y + 4 = 3x$$

Divide by 3:

$$x = \frac{5y + 4}{3}$$

Answer: $x = \frac{5y + 4}{3}$

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Example 7:

$$T = \frac{2πr}{v}$$

Make v the subject.

Solution:

$$T = \frac{2πr}{v}$$

Multiply both sides by $\mathrm{v:}$

$$Tv = 2πr$$

Divide both sides by $T$:

$$v = \frac{2πr}{T}$$

Answer: $v = \frac{2πr}{T}$

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Example 8:

$$I = \frac{V}{R}$$

Make R the subject.

Solution:

$$I = \frac{V}{R}$$

Multiply both sides by $\mathrm{R:}$

$$IR = V$$

Divide both sides by $\mathrm{I:}$

$$R = \frac{V}{I}$$

Answer: $R = \frac{V}{I}$

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Example 9:

$$p = \frac{2x + 3y}{4}$$

Make y the subject.

Solution:

$$4p = 2x + 3y$$

Subtract $2x$ from both sides:

$$4p - 2x = 3y$$

Divide both sides by 3:

$$y = \frac{4p - 2x}{3}$$

Answer: $y = \frac{4p - 2x}{3}$

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Example 10:

$$x=\frac{2y+1}{y-3}$$

Make y the subject.

Solution:
Multiply both sides by $(y-3):$

$$x(y - 3) = 2y + 1$$

Expand:

$$xy - 3x = 2y + 1$$

Group all $y$ terms on one side:

$$xy-2y=3x+1$$

Factor out y:

$$y(x - 2) = 3x + 1$$

Divide by (x - 2):

$$y = \frac{3x + 1}{x - 2}$$

Answer: $y = \frac{3x + 1}{x - 2}$​

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Example 11:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

Make v the subject.

Solution:
Subtract $\frac{1}{u}$ from both sides:

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$

Simplify the right-hand side:

$$\frac{1}{v} = \frac{u - f}{fu}$$

Take reciprocals:

$$v=\frac{fu}{u-\mathrm{f}}$$

Answer: $v=\frac{fu}{u-\mathrm{f}}$

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Example 12:

$$C=\frac{5}{9}(F-32)$$

Make F the subject.

Solution:
Multiply both sides by $\frac{9}{5}$:

$$\frac{9C}{5} = F - 32$$

Add 32 to both sides:

$$F=\frac{9C}{5}+32$$

Answer: $F = \frac{9C}{5} + 32$

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Example 13 (Advanced):

$$y = \frac{x - 1}{x + 2}$$

Make x the subject.

Solution:
Multiply both sides by $(x+2):$

$$y(x + 2) = x - 1$$

Expand:

$$yx + 2y = x - 1$$

Bring $\mathrm{x-terms\; together:}$

$$yx - x = -1 - 2y$$

Factor out $x$:

$$x(y - 1) = -1 - 2y$$

Divide by $(y-1)$:

$$x = \frac{-1 - 2y}{y - 1}$$

Answer: $x = \frac{-1 - 2y}{y - 1}$

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          Tips for Changing the Subject

• Treat all letters like numbers - apply the same algebraic operations.

• When a variable appears more than once, factorize to group it together.

• Use cross multiplication carefully when fractions are involved.

• Always check your final answer by substituting values to confirm the equality still holds.

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Practice Questions

Try these on your own:

1. $y = 3x - 7$ → make $x$ the subject.

2. $s = ut + \frac{1}{2}at^2$ → make $t$ the subject.

3. $m = \frac{2p + q}{r}$ → make $q$ the subject.

4. $E = \frac{mc^2}{2r}$ → make $m$ the subject.

5. $P = \frac{2L + 2B}{3H}$ → make $\mathrm{H\; the\; subject.}$

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Conclusion

Once you master it, you can handle physics, chemistry, and math formulas with confidence. So try your hands on as many examples as you can. All the best and see you on  another topic.