How do you simplify $\dfrac{3}{{{y^2} - 3y + 2}} + \dfrac{5}{{{y^2} - 1}}$?

Hint:In this question, we need to simplify the given fractions. Firstly, we simplify the denominator of the first term. We find the factors using a factoring method. Then we simplify the denominator of the second term using the difference of square formula. After that we substitute the obtained expression in the denominator of the first and second term. Then we can find a common denominator for both the terms and then we simplify further. We then use the distributive property to multiply the terms in the numerator. At last we will get a required result in the simplified form.

Complete step by step answer:
Given $\dfrac{3}{{{y^2} - 3y + 2}} + \dfrac{5}{{{y^2} - 1}}$ …… (1)
We are asked to simplify the above expression given in the equation (1).
Firstly, we will factor the denominator of the first term.
Consider the denominator ${y^2} - 3y + 2$
Note that this is a quadratic equation of degree 2.
Consider an equation of the form $a{y^2} + by + c$, where a, b, c are any real numbers.
We rewrite the middle term as a sum of two terms in such a way that their product is $a \cdot c$ and their sum is b.
In the given equation we have $a = 1$, $b = - 3$ and $c = 2$
We split the middle term $ - 3y$ as, $ - 3y = - 2y - y$.
Note that their product is,
 $a \cdot c = 1 \times (2)$
$ \Rightarrow a \cdot c = 2$
$ \Rightarrow 2 = ( - 2)( - 1)$.
Note that their sum is,
$b = - 3$
$ \Rightarrow - 3 = - 2 - 1$.
Hence the equation${y^2} - 3y + 2$ can be written as,
$ \Rightarrow {y^2} - 2y - y + 2$
Factor out the greatest common factor from each group, we get,
$ \Rightarrow y(y - 2) - 1(y - 2)$
Now factor the polynomial by factoring out the greatest common factor $y - 2$, we get,
$ \Rightarrow (y - 2)(y - 1)$
Hence ${y^2} - 3y + 2 = (y - 2)(y - 1)$ …… (2)
Consider the denominator of the second term given by ${y^2} - 1$.
This can be written as, ${y^2} - {1^2}$
Since both the terms are perfect squares, we factor it using the difference of squares formula given by, ${a^2} - {b^2} = (a - b)(a + b)$
Here $a = y$ and $b = 1$.
Hence we get, ${y^2} - 1 = (y - 1)(y + 1)$ …… (3)
Substituting the equation (2) and (3) in the equation (1), we get,
$ \Rightarrow \dfrac{3}{{(y - 2)(y - 1)}} + \dfrac{5}{{(y - 1)(y + 1)}}$
Take the common denominator as $(y - 2)(y - 1)(y + 1)$ and simplify the expression.
$ \Rightarrow \dfrac{{3(y + 1) + 5(y - 2)}}{{(y - 2)(y - 1)(y + 1)}}$
Now we multiply the each pair of terms in the numerator by using the distributive property which is given by,
$a \cdot (b + c) = ab + ac$
$a \cdot (b - c) = ab - ac$
Hence we get,
$ \Rightarrow \dfrac{{3y + 3 \times 1 + 5y - 5 \times 2}}{{(y - 2)(y - 1)(y + 1)}}$
Simplifying the above expression we get,
$ \Rightarrow \dfrac{{3y + 3 + 5y - 10}}{{(y - 2)(y - 1)(y + 1)}}$
Combining the like terms and solving we get,
$ \Rightarrow \dfrac{{8y - 7}}{{(y - 2)(y - 1)(y + 1)}}$

Hence the simplified form of $\dfrac{3}{{{y^2} - 3y + 2}} + \dfrac{5}{{{y^2} - 1}}$ is given by $\dfrac{{8y - 7}}{{(y - 2)(y - 1)(y + 1)}}$.

Note: Even though the given expression looks complicated, but it is simple to solve. Students must simplify the denominator of the fractions given and find the common denominator. So that it becomes easier to solve and simplify further.
Students must know the following facts to solve such problems.
(1) Consider an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
We rewrite the middle term as a sum of two terms in such a way that their product is $a \cdot c$ and their sum is b, to find the factors.
(2) The difference of square formula given by, ${a^2} - {b^2} = (a - b)(a + b)$
(3) The distributive property of addition and subtraction given by,
$a \cdot (b + c) = ab + ac$
$a \cdot (b - c) = ab - ac$