Question

practice: subtract two rational expressions
\\(\frac{4}{x + 1} - \frac{3}{2x - 3}\\)
which of the following is equivalent to the expression above?
choose 1 answer:
a \\(\frac{1}{4 - x}\\)
b \\(\frac{1}{2x^2 - x - 3}\\)
c \\(\frac{5x - 4}{2x^2 - x - 3}\\)
d \\(\frac{5(x - 3)}{2x^2 - x - 3}\\)

Explanation:

Step1: Find a common denominator

The denominators are \(x + 1\) and \(2x - 3\). The common denominator is \((x + 1)(2x - 3)\).
Rewrite each fraction with the common denominator:
\(\frac{4}{x + 1}=\frac{4(2x - 3)}{(x + 1)(2x - 3)}\)
\(\frac{3}{2x - 3}=\frac{3(x + 1)}{(x + 1)(2x - 3)}\)

Step2: Subtract the fractions

\(\frac{4(2x - 3)}{(x + 1)(2x - 3)}-\frac{3(x + 1)}{(x + 1)(2x - 3)}=\frac{4(2x - 3)-3(x + 1)}{(x + 1)(2x - 3)}\)

Step3: Expand and simplify the numerator

Expand the numerator:
\(4(2x - 3)-3(x + 1)=8x - 12 - 3x - 3\)
Combine like terms:
\(8x - 3x - 12 - 3 = 5x - 15\)? Wait, no, wait: \(8x - 3x = 5x\), \(-12 - 3=-15\)? Wait, no, let's recalculate:
Wait, \(4(2x - 3)=8x - 12\), \(3(x + 1)=3x + 3\), so subtracting: \(8x - 12-(3x + 3)=8x - 12 - 3x - 3 = 5x - 15\)? Wait, but the options have \(5x - 4\) or others. Wait, maybe I made a mistake. Wait, the original problem is \(\frac{4}{x + 1}-\frac{3}{2x - 3}\). Let's re - do the numerator:

\(4(2x - 3)-3(x + 1)=8x-12 - 3x - 3=5x-15\)? Wait, that's not matching the options. Wait, maybe I misread the problem. Wait, the problem is \(\frac{4}{x + 1}-\frac{3}{2x - 3}\). Wait, let's check the options again. Wait, maybe I made a mistake in expansion.

Wait, \(4(2x - 3)=8x - 12\), \(3(x + 1)=3x + 3\). Then \(8x-12-(3x + 3)=8x - 12 - 3x - 3 = 5x-15\). But the denominator is \((x + 1)(2x - 3)=2x^{2}-3x + 2x-3=2x^{2}-x - 3\).

Wait, but option C is \(\frac{5x - 4}{2x^{2}-x - 3}\), option D is \(\frac{5(x - 3)}{2x^{2}-x - 3}=\frac{5x-15}{2x^{2}-x - 3}\). Wait, maybe the original problem was \(\frac{4}{x + 1}-\frac{3}{2x + 3}\)? No, the problem says \(2x - 3\). Wait, maybe I made a mistake. Wait, let's check the problem again.

Wait, the user's problem: \(\frac{4}{x + 1}-\frac{3}{2x - 3}\). Let's re - calculate the numerator:

\(4(2x - 3)-3(x + 1)=8x-12-3x - 3 = 5x-15=5(x - 3)\). And the denominator is \((x + 1)(2x - 3)=2x^{2}-3x+2x - 3=2x^{2}-x - 3\). So the expression is \(\frac{5(x - 3)}{2x^{2}-x - 3}\), which is option D? Wait, but let's check the options again.

Wait, option D is \(\frac{5(x - 3)}{2x^{2}-x - 3}\), option C is \(\frac{5x - 4}{2x^{2}-x - 3}\). Wait, maybe I misread the numerator. Wait, maybe the first term is \(\frac{4}{x + 1}\) and the second term is \(\frac{3}{2x+3}\)? No, the problem says \(2x - 3\). Wait, let's check the arithmetic again:

\(4(2x - 3)=8x - 12\)

\(3(x + 1)=3x + 3\)

\(8x-12-(3x + 3)=8x-12 - 3x - 3=5x-15 = 5(x - 3)\)

Denominator: \((x + 1)(2x - 3)=2x^{2}-3x+2x - 3=2x^{2}-x - 3\)

So the result is \(\frac{5(x - 3)}{2x^{2}-x - 3}\), which is option D. Wait, but let's check the options again. The options are:

A. \(\frac{1}{4 - x}\)

B. \(\frac{1}{2x^{2}-x - 3}\)

C. \(\frac{5x - 4}{2x^{2}-x - 3}\)

D. \(\frac{5(x - 3)}{2x^{2}-x - 3}\)

Wait, maybe I made a mistake in the sign. Wait, the original expression is \(\frac{4}{x + 1}-\frac{3}{2x - 3}\), so it's \(4(2x - 3)-3(x + 1)\). Let's recalculate:

\(4(2x-3)=8x - 12\)

\(3(x + 1)=3x + 3\)

\(8x-12-3x - 3=5x-15 = 5(x - 3)\)

Yes, so the numerator is \(5(x - 3)\) and the denominator is \(2x^{2}-x - 3\), so the answer is D.

Wait, but let's check again. Wait, maybe the problem was \(\frac{4}{x + 1}-\frac{3}{2x+3}\)? No, the problem says \(2x - 3\). Alternatively, maybe I misread the numerator of the options. Wait, option D is \(\frac{5(x - 3)}{2x^{2}-x - 3}\), which is equal to \(\frac{5x-15}{2x^{2}-x - 3}\), and our numerator is \(5x - 15\), so that's correct.

Answer:

D. \(\frac{5(x - 3)}{2x^{2}-x - 3}\)