Question

adding / subtracting rational expressions (lev
question
perform the operation and combine to one fraction.
\\(\dfrac{5}{x^2 - 4x + 3} - \dfrac{2x}{x - 1}\\)
answer
attempt 1 out of 2

Explanation:

Step1: Factor the denominator

Factor \(x^2 - 4x + 3\). We need two numbers that multiply to 3 and add to -4, which are -1 and -3. So, \(x^2 - 4x + 3=(x - 1)(x - 3)\). The expression becomes \(\frac{5}{(x - 1)(x - 3)}-\frac{2x}{x - 1}\).

Step2: Find the common denominator

The common denominator of \((x - 1)(x - 3)\) and \(x - 1\) is \((x - 1)(x - 3)\). Rewrite the second fraction with the common denominator: \(\frac{2x}{x - 1}=\frac{2x(x - 3)}{(x - 1)(x - 3)}\).

Step3: Subtract the fractions

Now subtract the two fractions: \(\frac{5}{(x - 1)(x - 3)}-\frac{2x(x - 3)}{(x - 1)(x - 3)}=\frac{5-2x(x - 3)}{(x - 1)(x - 3)}\).

Step4: Simplify the numerator

Expand the numerator: \(5-2x(x - 3)=5-2x^2 + 6x=-2x^2+6x + 5\). So the fraction is \(\frac{-2x^2 + 6x + 5}{(x - 1)(x - 3)}\) (or we can also factor out a negative sign from the numerator: \(\frac{-(2x^2 - 6x - 5)}{(x - 1)(x - 3)}\), but the first form is also correct). Alternatively, if we expand and simplify:
Wait, let's re - check the subtraction:
\(\frac{5-(2x^{2}-6x)}{(x - 1)(x - 3)}=\frac{5 - 2x^{2}+6x}{(x - 1)(x - 3)}=\frac{-2x^{2}+6x + 5}{(x^{2}-4x + 3)}\) (since \((x - 1)(x - 3)=x^{2}-4x + 3\))

Answer:

\(\frac{-2x^{2}+6x + 5}{x^{2}-4x + 3}\) (or \(\frac{-2x^{2}+6x + 5}{(x - 1)(x - 3)}\))