Question

1. simplify the following rational expressions.

a. \\(\frac{5}{x^2 + 3} + \frac{7}{2x^2 + 1}\\)
b. \\(\frac{4x}{x + 5} - \frac{x - 3}{2x - 7}\\)
c. \\(\frac{3x^2}{x^2 - 3} - \frac{5}{2x^2 + 1}\\)

Explanation:

Part a

Step1: Find the common denominator

The denominators are \(x^2 + 3\) and \(2x^2+1\). The common denominator is \((x^2 + 3)(2x^2 + 1)\).

Step2: Rewrite each fraction with the common denominator

\(\frac{5}{x^2 + 3}=\frac{5(2x^2 + 1)}{(x^2 + 3)(2x^2 + 1)}\) and \(\frac{7}{2x^2 + 1}=\frac{7(x^2 + 3)}{(x^2 + 3)(2x^2 + 1)}\)

Step3: Add the numerators

\[

$$\begin{align*} &\frac{5(2x^2 + 1)+7(x^2 + 3)}{(x^2 + 3)(2x^2 + 1)}\\ =&\frac{10x^2+5 + 7x^2+21}{(x^2 + 3)(2x^2 + 1)}\\ =&\frac{17x^2 + 26}{(x^2 + 3)(2x^2 + 1)} \end{align*}$$

\]

Step1: Find the common denominator

The denominators are \(x + 5\) and \(2x-7\). The common denominator is \((x + 5)(2x-7)\).

Step2: Rewrite each fraction with the common denominator

\(\frac{4x}{x + 5}=\frac{4x(2x - 7)}{(x + 5)(2x-7)}\) and \(\frac{x - 3}{2x-7}=\frac{(x - 3)(x + 5)}{(x + 5)(2x-7)}\)

Step3: Subtract the numerators

\[

$$\begin{align*} &\frac{4x(2x - 7)-(x - 3)(x + 5)}{(x + 5)(2x-7)}\\ =&\frac{8x^2-28x-(x^2 + 5x-3x - 15)}{(x + 5)(2x-7)}\\ =&\frac{8x^2-28x-(x^2 + 2x - 15)}{(x + 5)(2x-7)}\\ =&\frac{8x^2-28x - x^2-2x + 15}{(x + 5)(2x-7)}\\ =&\frac{7x^2-30x + 15}{(x + 5)(2x-7)} \end{align*}$$

\]

Step1: Find the common denominator

The denominators are \(x^2-3\) and \(2x^2 + 1\). The common denominator is \((x^2-3)(2x^2 + 1)\).

Step2: Rewrite each fraction with the common denominator

\(\frac{3x^2}{x^2-3}=\frac{3x^2(2x^2 + 1)}{(x^2-3)(2x^2 + 1)}\) and \(\frac{5}{2x^2 + 1}=\frac{5(x^2-3)}{(x^2-3)(2x^2 + 1)}\)

Step3: Subtract the numerators

\[

$$\begin{align*} &\frac{3x^2(2x^2 + 1)-5(x^2-3)}{(x^2-3)(2x^2 + 1)}\\ =&\frac{6x^4+3x^2-5x^2 + 15}{(x^2-3)(2x^2 + 1)}\\ =&\frac{6x^4-2x^2 + 15}{(x^2-3)(2x^2 + 1)} \end{align*}$$

\]

Answer:

\(\frac{17x^2 + 26}{(x^2 + 3)(2x^2 + 1)}\)

Part b