Question

adding / subtracting rational expressions (lev
question
perform the operation and combine to one
fraction.
\\(\dfrac{2x}{x + 5} - \dfrac{3x + 2}{x^2 + 7x + 10}\\)
answer
attempt 1 out of 2
submit answer

Explanation:

Step1: Factor the denominator

Factor \(x^2 + 7x + 10\). We need two numbers that multiply to 10 and add to 7. Those numbers are 2 and 5. So, \(x^2 + 7x + 10=(x + 2)(x + 5)\).
The expression becomes \(\frac{2x}{x + 5}-\frac{3x + 2}{(x + 2)(x + 5)}\).

Step2: Find a common denominator

The common denominator of \((x + 5)\) and \((x + 2)(x + 5)\) is \((x + 2)(x + 5)\). Rewrite the first fraction with the common denominator: \(\frac{2x(x + 2)}{(x + 2)(x + 5)}-\frac{3x + 2}{(x + 2)(x + 5)}\).

Step3: Expand the numerator of the first fraction

Expand \(2x(x + 2)\): \(2x(x + 2)=2x^2 + 4x\). Now the expression is \(\frac{2x^2 + 4x-(3x + 2)}{(x + 2)(x + 5)}\).

Step4: Simplify the numerator

Distribute the negative sign: \(2x^2 + 4x - 3x - 2=2x^2 + x - 2\). Wait, let's check that again. Wait, \(2x^2+4x - 3x - 2 = 2x^2+x - 2\)? Wait, no, wait, maybe I made a mistake. Wait, let's redo the numerator:

\(2x(x + 2)-(3x + 2)=2x^2 + 4x - 3x - 2=2x^2 + x - 2\)? Wait, but let's factor \(2x^2 + x - 2\)? Wait, maybe I made a mistake in the expansion. Wait, no, let's check the original problem again. Wait, the first term is \(\frac{2x}{x + 5}\), the second is \(\frac{3x + 2}{x^2 + 7x + 10}=\frac{3x + 2}{(x + 2)(x + 5)}\). So common denominator is \((x + 2)(x + 5)\), so \(\frac{2x(x + 2)}{(x + 2)(x + 5)}-\frac{3x + 2}{(x + 2)(x + 5)}=\frac{2x(x + 2)-(3x + 2)}{(x + 2)(x + 5)}\).

Now expand \(2x(x + 2)=2x^2 + 4x\), then subtract \(3x + 2\): \(2x^2 + 4x - 3x - 2=2x^2 + x - 2\). Wait, but maybe I made a mistake here. Wait, let's check if \(2x^2 + x - 2\) can be factored? Let's use the quadratic formula: \(x=\frac{-1\pm\sqrt{1 + 16}}{4}=\frac{-1\pm\sqrt{17}}{4}\), which are not integers. So maybe my approach is wrong. Wait, no, wait, maybe I messed up the sign. Wait, no, let's check again.

Wait, maybe I made a mistake in the factoring. Wait, \(x^2 + 7x + 10=(x + 2)(x + 5)\), that's correct. Then common denominator is \((x + 2)(x + 5)\), so \(\frac{2x(x + 2)}{(x + 2)(x + 5)}-\frac{3x + 2}{(x + 2)(x + 5)}\). Then numerator: \(2x(x + 2)-(3x + 2)=2x^2 + 4x - 3x - 2=2x^2 + x - 2\). Wait, but maybe the numerator can be simplified further? Wait, no, maybe I made a mistake in the problem. Wait, let's check the original problem again. Wait, the problem is \(\frac{2x}{x + 5}-\frac{3x + 2}{x^2 + 7x + 10}\). Wait, maybe the numerator is \(2x(x + 2)-(3x + 2)=2x^2 + 4x - 3x - 2=2x^2 + x - 2\). But maybe I made a mistake in the sign. Wait, no, the subtraction is of the second fraction, so it's \(2x(x + 2)\) minus \((3x + 2)\). So that's correct. Wait, but maybe the numerator factors? Let's check \(2x^2 + x - 2\). The discriminant is \(1^2 - 4\times2\times(-2)=1 + 16 = 17\), which is not a perfect square, so it doesn't factor nicely. Wait, but maybe I made a mistake in the expansion. Wait, no, \(2x(x + 2)=2x^2 + 4x\), then subtract \(3x + 2\), so \(2x^2 + 4x - 3x - 2=2x^2 + x - 2\). So the final fraction is \(\frac{2x^2 + x - 2}{(x + 2)(x + 5)}\)? Wait, but that seems complicated. Wait, maybe I made a mistake in the problem. Wait, let's check again. Wait, maybe the original problem was \(\frac{2x}{x + 5}-\frac{3x - 2}{x^2 + 7x + 10}\)? No, the user wrote \(3x + 2\). Wait, maybe I made a mistake in the sign. Wait, no, the problem is \(\frac{2x}{x + 5}-\frac{3x + 2}{x^2 + 7x + 10}\). So the numerator is \(2x(x + 2)-(3x + 2)=2x^2 + 4x - 3x - 2=2x^2 + x - 2\). So the simplified form is \(\frac{2x^2 + x - 2}{(x + 2)(x + 5)}\) or we can leave it as \(\frac{2x^2 + x - 2}{x^2 + 7x + 10}\) (since \((x + 2)(x + 5)=x^2 + 7x + 10\)). Wait, but maybe I made a mistake. Wait, let's check with another approach. Wait, maybe the numerator is \(2x(x + 2)-(3x + 2)=2x^2 + 4x - 3x - 2=2x^2 + x - 2\). So that's the numerator. So the final answer is \(\frac{2x^2 + x - 2}{(x + 2)(x + 5)}\) or \(\frac{2x^2 + x - 2}{x^2 + 7x + 10}\). Wait, but maybe I made a mistake in the expansion. Wait, no, let's check with x=0. Let's plug x=0 into the original expression: \(\frac{0}{0 + 5}-\frac{0 + 2}{0 + 0 + 10}=0-\frac{2}{10}=-\frac{1}{5}\). Now plug x=0 into the final expression: \(\frac{0 + 0 - 2}{(0 + 2)(0 + 5)}=\frac{-2}{10}=-\frac{1}{5}\). So that works. So the answer is correct.

Answer:

\(\frac{2x^2 + x - 2}{(x + 2)(x + 5)}\) (or \(\frac{2x^2 + x - 2}{x^2 + 7x + 10}\))