DB125: Maths for Diploma Studies: Rearranging Equations – When the Formula has Two Terms that Contain the Subject we need to Isolate Reinforcement Questions

Rearranging Equations – When the Formula has Two Terms that Contain the Subject we need to Isolate Reinforcement Questions

Rearrange the equation below to make x the subject:

$begin mathsize 14px style a left parenthesis x minus space 5 right parenthesis space equals space 3 left parenthesis 2 space minus space 7 x right parenthesis end style$

First, we expand the brackets by multiplying the terms in the brackets by the term outside, yielding:

$begin mathsize 14px style a x space minus space space 5 a space equals space 6 space minus space 21 x end style$

Now we need to bring all of the x terms to the same side of the equation by adding 21x to both sides:

$begin mathsize 14px style a x space plus space 21 x space minus space 5 a space equals space 6 end style$

Next, move all of the non x terms to the other side of the equation by adding 5a to both sides:

$begin mathsize 14px style a x space plus space 21 x space equals space 6 space plus space 5 a end style$

Then we factorise the left-hand side of the equation by x:

$begin mathsize 14px style x left parenthesis a space plus space 21 right parenthesis space equals space 6 space plus space 5 a end style$

We then divide both sides of the equation by the bracketed items next to x:

$begin mathsize 14px style x space equals space fraction numerator 6 space plus space 5 a over denominator a space plus space 21 end fraction end style$

Isolate m in this equation:

$begin mathsize 14px style F space equals space fraction numerator m v space minus space m u over denominator t end fraction end style$

First, we multiply both sides of the equation by t to get:

$begin mathsize 14px style F t space equals space m v space minus space m u end style$

Then factorise by m:

$begin mathsize 14px style F t space equals space m left parenthesis v space minus space u right parenthesis end style$

Next, we make m, the subject of the equation, by dividing both sides by v - u:

$begin mathsize 14px style m space equals space fraction numerator F t over denominator v space minus space u end fraction end style$

Isolate b in the following equation:

$begin mathsize 14px style a b space minus space c space equals space 3 b space plus space d end style$

We would first bring all of the b terms to the same side by subtracting 3b from both sides:

$begin mathsize 14px style a b space minus space 3 b space minus space c space equals space d end style$

Then we move all non-b terms to the other side of the equation by adding c to both sides:

$begin mathsize 14px style a b space minus space 3 b space equals space d space plus space c end style$

Next, we factorise the left-hand side of the equation by b:

$begin mathsize 14px style b left parenthesis a space minus space 3 right parenthesis space equals space d space plus space c end style$

Finally, we divide both sides of the equation by the bracketed terms adjacent to b:

$begin mathsize 14px style b space equals space fraction numerator d space plus space c over denominator a space minus space 3 end fraction end style$

Make x the subject in the following equation:

$begin mathsize 14px style 2 left parenthesis x y space plus space z right parenthesis space equals space 5 left parenthesis x space minus space z right parenthesis end style$

We start by expanding both brackets by multiplying the terms inside by the terms outside:

$begin mathsize 14px style 2 x y space plus space 2 z space equals space 5 x space minus space 5 z end style$

We need to make $begin mathsize 14px style 5 x end style$ active in the equation by changing how it is written:

$begin mathsize 14px style 2 x y space plus space 2 z space equals space minus 5 z space plus space space 5 x end style$

Now we move all x terms to the left-hand side of the equation by subtracting $begin mathsize 14px style 5 x end style$ from both sides:

$begin mathsize 14px style 2 x y space minus space 5 x space plus space 2 z space equals negative 5 z end style$

Then we move all non-x terms to the right-hand side of the equation by subtracting $begin mathsize 14px style 2 z end style$ from both sides:

$begin mathsize 14px style 2 x y space minus space 5 x space equals space minus 5 z space minus space 2 z end style$

Now we factorise the left-hand side of the equation by x:

$begin mathsize 14px style x left parenthesis 2 y space minus space 5 right parenthesis space equals space minus 5 z space minus space 2 z end style$

And then we can simplify the right-hand side of the equation:

$begin mathsize 14px style x left parenthesis 2 y space minus space 5 right parenthesis space equals space minus 7 z end style$

Finally, we divide both sides of the equation by the bracketed items adjacent to the x term:

$begin mathsize 14px style x space equals space fraction numerator negative 7 z over denominator 2 y space minus space 5 end fraction end style$

Rearrange this equation to make b the subject:

$begin mathsize 14px style P space equals space fraction numerator a b space plus space c over denominator b space minus space d end fraction end style$

First, let’s get rid of the fraction by multiplying both sides by the denominator:

$begin mathsize 14px style P left parenthesis b space minus space d space right parenthesis space equals space a b space plus space c end style$

Then, expand the brackets by multiplying by P:

$begin mathsize 14px style P b space minus space P d space equals space a b space plus space c end style$

Now we bring all of the b terms to the same side of the equation by subtracting ab from both sides:

$begin mathsize 14px style P b space minus space a b space minus space P d space equals space c end style$

Next, we factorise by b:

$begin mathsize 14px style b left parenthesis P space minus space a right parenthesis space minus space P d space equals space c end style$

Now we can move Pd across the equation by adding it to both sides:

$begin mathsize 14px style b left parenthesis P space minus space a right parenthesis space equals space c space plus space P d end style$

And finally, we can divide both sides of the equation by $begin mathsize 14px style P space minus space a end style$ :

$begin mathsize 14px style b space equals space fraction numerator c space plus space P d over denominator P space minus space a end fraction end style$

In the next section, will we learn what to do when a term appears on both sides of the equation.

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Maths for Diploma Studies

Rearranging Equations – When the Formula has Two Terms that Contain the Subject we need to Isolate Reinforcement Questions

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