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I am looking at a math question that has simplified this:

enter image description here

into this:

enter image description here

Can somebody explain the process for how this simplification was made? i.e. how does the denominator get broken down to those two terms?

Startec
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2 Answers2

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Here's a quick and easy way for simple or basic fractions$$\frac 1{y(1-y)}=\frac {\color{blue}{1-y}+y}{y(\color{blue}{1-y})}=\frac {1-y}{y(1-y)}+\frac y{y(1-y)}=\frac 1y+\frac 1{1-y}\color{brown}{=\frac 1y-\frac 1{y-1}}$$

Frank W
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We write $$ \frac {1}{y(1-y)}=\frac {A}{y}+\frac {B}{1-y}$$

Upon taking common denominator , we get $$\frac {A(1-y)+By}{Y(1-y)}=\frac {1}{y(1-y)}$$

Therefore we want to have,$$ A(1-y)+B(y)=1$$

Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$

Thus $$ \frac {1}{y(1-y)}=\frac {1}{y}+\frac {1}{1-y}=\frac {1}{y}-\frac {1} {y-1} $$

Mohammad Riazi-Kermani
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  • Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1" – Startec Dec 27 '18 at 07:48
  • We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$ – Mohammad Riazi-Kermani Dec 27 '18 at 09:50