How do you solve \[\dfrac{2}{b-2}=\dfrac{b}{{{b}^{2}}-3b+2}+\dfrac{b}{2b-2}\]?

Hint: To solve this equation first we have to make the denominators the same.
We know that if we want to add the fractions directly or we want to add the numerators of the fractions directly then the denominator should be the same as the fractions.
For example
\[\dfrac{a}{4b+2}+\dfrac{b}{2b+1}\]
So,
To add the numerators of these fractions we have to make the denominators equal
In the fraction \[\dfrac{b}{2b+1}\] the denominator is half of the denominator of \[\dfrac{a}{4b+2}\] so by multiplying the numerator and denominator of \[\dfrac{b}{2b+1}\] the denominator of both fractions will get equal
\[\dfrac{a}{4b+2}+\dfrac{2b}{2\left( 2b+1 \right)}=\dfrac{a}{4b+2}+\dfrac{2b}{4b+2}\]
\[=\dfrac{a+2b}{4b+2}\]
This is how fractions can be added.

Complete step by step solution:
We are looking for common denominators so we need to make them all the same format type which means we have to make the denominators the same.
Consider \[{{b}^{2}}-3b+2\]
This is a quadratic equation.
So factors are two and that is 1 and 2.
Note that
\[\left( -1 \right)\times \left( -2 \right)=+2\] and \[-1-2=-3\]
So we have
\[{{b}^{2}}-3b+2=\left( b-1 \right)\left( b-2 \right)\]
Rewrite the given equation as
\[\dfrac{2}{b-2}=\dfrac{b}{\left( b-1 \right)\left( b-2 \right)}+\dfrac{b}{2b-2}\]
Now taking \[\left( 2b-2 \right)\] we can write it as \[2\left( b-1 \right)\] so, we now have
\[\dfrac{2}{b-2}=\dfrac{b}{\left( b-1 \right)\left( b-2 \right)}+\dfrac{b}{2\left( b-1 \right)}\]
Now from here we can see that the denominators will be equal it we make all the denominators as \[2\left( b-1 \right)\left( b-2 \right)\]
Multiplying each part of the equation by 1.
Therefore,
\[\left[ \dfrac{2}{b-2}\times 1 \right]=\left[ \dfrac{b}{\left( b-1 \right)\left( b-2 \right)}\times 1 \right]+\left[ \dfrac{b}{2\left( b-1 \right)}\times 1 \right]\]
What follows may be folded if the equation is wider than the page.
\[\left[ \dfrac{2}{b-2}\times \dfrac{2\left( b-1 \right)}{2\left( b-1 \right)} \right]=\left[ \dfrac{b}{\left( b-1 \right)\left( b-2 \right)}\times \dfrac{2}{2} \right]+\left[ \dfrac{b}{2\left( b-1 \right)}\times \dfrac{b-2}{b-2} \right]\]
\[\dfrac{4\left( b-1 \right)}{2\left( b-1 \right)\left( b-2 \right)}=\dfrac{2b}{2\left( b-1 \right)\left( b-2 \right)}+\dfrac{b\left( b-2 \right)}{2\left( b-1 \right)\left( b-2 \right)}\]
Now, multiplying both sides by \[2\left( b-1 \right)\left( b-2 \right)\] giving:
\[4\left( b-1 \right)=2b+b(b-2)\]
\[4b-4=2b+{{b}^{2}}-2b\]
\[{{b}^{2}}-4b+4=0\]
Now, we have got a quadratic equation.
We will solve it as standard quadratic equation
\[b=\dfrac{+4\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( +4 \right)}}{2\left( 1 \right)}\]
\[b=2\pm \dfrac{\sqrt{0}}{2}\]
\[b=2\]
Now let's check.
LHS \[\dfrac{2}{b-2}=\dfrac{2}{2-2}=\dfrac{2}{0}\] undefined
RHS \[\dfrac{b}{\left( b-1 \right)\left( b-2 \right)}+\dfrac{b}{2b-2}\]
\[=\dfrac{2}{\left( 2-1 \right)\left( 2-2 \right)}+\dfrac{2}{2}\]
\[=\dfrac{2}{0}+1\]
\[\dfrac{2}{0}\] undefined.
Both LHS and RHS are undefined.

Note: When you get the value of \[b\] always check whether the value is correct or not by putting the value of \[b\] in the given equation if both RHS and LHS is equal then the value if correct.