Question

# We need to calculate the simplified form of the expression, frac{c^{2} + 13c + 18}{c^{2} - 9} + frac{c + 1}{c + 3} - frac{c + 8}{c - 3}

We need to calculate the simplified form of the expression,
$$\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}$$

2021-01-23
Given Information:
The given expression is $$\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}$$
Formula:
The algebraic identity used is $$a^{2}\ -\ b^{2}=(a\ +\ b)(a\ -\ b).$$
Calculation:
Consider the expression, $$\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}$$
First, factor the denominators of this expression and apply
$$a^{2}\ -\ b^{2}=(a\ +\ b)(a\ -\ b)\ in\ c^{2}\ -\ 9$$ as follows:
$$\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}=\frac{c^{2}\ +\ 13c\ +\ 18}{(c\ -\ 3)(c\ +\ 3)}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}$$
Now, consider the denominators, $$(c\ -\ 3)(c\ +\ 3),\ c\ +\ 3\ and\ c\ -\ 3.$$
The least common denominator $$(LCD)\ is\ (c\ -\ 3)(c\ +\ 3).$$
Now multiply the numerator and the denominator of $$\frac{c^{2}\ +\ 13c\ +\ 18}{(c\ -\ 3)(c\ +\ 3)}\ with\ 1,\ \frac{c\ +\ 1}{c\ +\ 3}\ with\ c\ -\ 3\ and\ \frac{c\ +\ 8}{c\ -\ 3}\ with\ c\ +\ 3$$ to get the LCD in the denominator as follows:
$$\frac{c^{2}\ +\ 13c\ +\ 18}{(c\ -\ 3)(c\ +\ 3)}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}=\ \frac{(c^{2}\ +\ 13c\ +\ 18)(1)}{(c\ -\ 3)(c\ +\ 3)(1)}\ +\ \frac{(c\ +\ 1)(c\ -\ 3)}{(c\ +\ 3)(c\ -\ 3)}\ -\ \frac{(c\ +\ 8)(c\ +\ 3)}{(c\ -\ 3)(c\ +\ 3)}$$
$$=\frac{(c^{2}\ +\ 13c\ +\ 18)(1)\ +\ (c\ +\ 1)(c\ -\ 3)\ -\ (c\ +\ 8)(c\ +\ 3)}{(c\ -\ 3)(c\ +\ 3)}$$
$$=\frac{c^{2}\ +\ 13c\ +\ 18\ +\ c^{2}\ -\ 3c\ +\ c\ -\ 3\ -\ c^{2}\ -\ 8c\ -\ 3c\ -\ 24}{(c\ -\ 3)(c\ +\ 3)}$$
Solve further, $$=\frac{c^{2}\ +\ 13c\ +\ 18\ +\ c^{2}\ -\ 3c\ +\ c\ -\ 3\ -\ c^{2}\ -\ 8c\ -\ 3c\ -\ 24}{(c\ -\ 3)(c\ +\ 3)}=\frac{c^{2}\ -\ 9}{c^{2}\ -\ 9}$$
$$= 1$$
Therefore, the simplified form of the expression,
$$\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}\ is\ 1$$