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}

Rational exponents and radicals
ANSWERED
asked 2021-01-22
We need to calculate the simplified form of the expression,
\(\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}\)

Answers (1)

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,
\(\frac{c^{2}\ +\ 13c\ +\ 18}{c^{2}\ -\ 9}\ +\ \frac{c\ +\ 1}{c\ +\ 3}\ -\ \frac{c\ +\ 8}{c\ -\ 3}\ is\ 1\)
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