CHANGE OF SUBJECT FORMULA

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The subject of a formula is the variable that is expressed in terms of the other variables.

In the relation $I=\frac{PRT}{100}$, $I$ is called the subject of the formula.

Sometimes, It may be necessary to change the subject of the formula or rearrange the relation in an equivalent form. A preponderance of mathematical formulas is derived using the change of subject of the formula. An instance of this is the "distance formula" which is calculated as:

$D=S\times T$

Where S is speed
            T is a Time.

But, do you know that we can derive the formula for speed by making speed the of the formula
$D=ST$.

Divide both sides by $T$

$\frac{D}{T}=\frac{S\require{cancel}\bcancel{T}}{\require{cancel}\bcancel{T}}$
$S=\frac{D}{T}$

The formula for time is also so derived using the change of subject formula.

The next 7 examples illustrate the change of subject formula

Example 1
Make $y$ the subject of the formula in the relation $x=3y$.

We simply divide both sides by $3$
$\frac{x}{3}=\frac{\require{cancel}\bcancel{3}y}{\require{cancel}\bcancel{3}}$
$\frac{x}{3}=y$
$y=\frac{x}{3}$

Example 2
Make $x$ the subject of the formula in the relation $C=a(x-1)-b(x-1)$.

$C=a(x-1)-b(x-1)$ can be rewritten as
$C=(a-b)(x-1)$

Divide both side by $(a-b)$
$\frac{C}{a-b}=\frac{\require{cancel}\bcancel{(a-b)}(x-1)}{\require{cancel}\bcancel{(a-b)}}$

$\frac{C}{a-b}=x-1$
$\frac{C}{a-b}+1=x$
$x=\frac{C}{a-b}+1$

Example 3
If $u=px+qx+c$, obtain a formula for $x$

Obtain a formula for $x$ is the same thing as to make $x$ the subject of the formula

Re-arrange so that all the variable with x will be at one side of the equation
$u-c=px+qx$

Factoring the R.H.S
$u-c=x(p+q)$

Divide both sides by $(p+q)$

$frac{(u-c)}{(p+q)}=\frac{x\require{cancel}\bcancel{(p+q)}}{\require{cancel}\bcancel{(p+q)}}$
$\frac{u-c}{p+q}=x$
$x=\frac{u-c}{p+q}$

Example 4
If $f=\frac{9x}{5}+32$, obtain a formula for $x$

Taking the L.C.M
$f=\frac{9x+160}{5}$

Cross multiply
$5\times f=9x+160$
$5f=9x+160$
$5f-160=9x$

Divide both side by $9$
$\frac{5f-160}{9}=\frac{9x}{9}$
$x=\frac{5f-160}{9}$

Example 5
Make $x$ the subject of the formula I'm the equation $a=\frac{b+cx}{d}$

Cross multiply
$ad=b+cx$
$ad-b=cx$

Divide both sides by $c$
$\frac{ad-b}{c}=\frac{cx}{c}$
$\frac{ad-b}{c}=x$
$x=\frac{ad-b}{c}$

Example 6
Make $x$ the subject of the formula
If $y=\frac{s}{x^2}-u$

Taking the L.C.M
$y=\frac{s-x^2u}{x^2}$

Cross multiply
$y\times x^2=s-x^2u$
$x^2y=s-x^2u$

Collect like term
$x^2y+x^2u=s$
$x^2(y+u)=s$

Divide both sides by $(y+u)$
$\frac{x^2\require{cancel}\bcancel{(y+u)}}{\require{cancel}\bcancel{(y+u)}}=\frac{s}{(y+u)}$
$x^2=\frac{s}{(y+u)}$

To get rid of the square root, we would square root both side, 


$(\sqrt{x})^2=\sqrt{\frac{s}{(y+u)}}$
$x=\sqrt{\frac{s}{(y+u)}}$

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Example 7
Obtain a formula for $q$ in the equation $v=\sqrt{u^2-2qs}$

To get rid of ths square root, you will square both side
$v^2=(\sqrt{u^2-2qs})^2$
$v^2=u^2-2qs$

Re-arrange so that all the variables with q will be at one side of the equation while the variable without q will be at the other side of the equation.

$2qs=u^2-v^2$
$q(2s)=u^2-v^2$

Divide both side by $2s$
$\frac{q\require{cancel}\bcancel{(2s)}}{\require{cancel}\bcancel{2s}}=\frac{u^2-v^2}{2s}$
$q=\frac{u^2-v^2}{2s}$
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