# ❊ Simplify : 1. $$\large{ \tt{ \frac{a + 2}{ {a}^{2} + a – 2 } + \frac{3}{ {a}^{2} – 1 } \: \: \{ANS : \frac{a + 4}{ {a}^{2} – 1 } ❊ Simplify : 1. \large{ \tt{ \frac{a + 2}{ {a}^{2} + a – 2 } + \frac{3}{ {a}^{2} – 1 } \: \: \{ANS : \frac{a + 4}{ {a}^{2} – 1 } \}}} 2. \large{ \tt{ \frac{1}{(a – b)(b – c)} + \frac{1}{(c – b)(a – c) } \: \: \{ANS : \frac{1}{(a – b)(a – c) } \}}} -Show your workings! * – Random/ Irrelevant answers will be reported!​ ### 0 thoughts on “❊ Simplify : 1. [tex] \large{ \tt{ \frac{a + 2}{ {a}^{2} + a – 2 } + \frac{3}{ {a}^{2} – 1 } \: \: \{ANS : \frac{a + 4}{ {a}^{2} – 1 }” 1. Answer: See Below. Step-by-step explanation: Problem 1) We want to simplify: [tex]\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$$

First, let’s factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$$

$$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$$

Factor:

$$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$$

Simplify:

$$\displaystyle =\frac{a+4}{(a-1)(a+1)}$$

We can expand. Therefore:

$$\displaystyle =\frac{a+4}{a^2-1}$$

Problem 2)

We want to simplify:

$$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$$

Again, let’s create a common denominator. First, let’s factor out a negative from the second term:

\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}

Now to create a common denominator, we can multiply the first term by (ac) and the second term by (ab). Hence:

$$\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}$$

Subtract the fractions:

$$\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}$$

Distribute and simplify:

$$\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}$$

Cancel. Hence:

$$\displaystyle =\frac{1}{(a-b)(a-c)}$$