Question

question
evaluate the limit: $limlimits_{x \to 9} \frac{2x - 2}{\frac{2}{3} - \frac{3x + 5}{x - 7}}$
answer attempt 1 out of 2
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Explanation:

Step1: Simplify the denominator

First, simplify the denominator \(\frac{2}{3}-\frac{3x + 5}{x - 7}\). Find a common denominator, which is \(3(x - 7)\).
\[

$$\begin{align*} \frac{2}{3}-\frac{3x + 5}{x - 7}&=\frac{2(x - 7)-3(3x + 5)}{3(x - 7)}\\ &=\frac{2x-14 - 9x - 15}{3(x - 7)}\\ &=\frac{-7x-29}{3(x - 7)} \end{align*}$$

\]

Step2: Rewrite the original limit

The original limit \(\lim_{x
ightarrow9}\frac{2x - 2}{\frac{2}{3}-\frac{3x + 5}{x - 7}}\) becomes \(\lim_{x
ightarrow9}\frac{2x - 2}{\frac{-7x - 29}{3(x - 7)}}\), which is equivalent to \(\lim_{x
ightarrow9}(2x - 2)\times\frac{3(x - 7)}{-7x - 29}\) (by dividing by a fraction is multiplying by its reciprocal).

Step3: Substitute \(x = 9\)

Now substitute \(x = 9\) into the expression:
For the numerator part: \(2(9)-2=18 - 2 = 16\), and \(3(9 - 7)=3\times2 = 6\).
For the denominator part: \(-7(9)-29=-63 - 29=-92\).
Then the expression is \(\frac{16\times6}{-92}=\frac{96}{-92}=-\frac{24}{23}\)

Answer:

\(-\frac{24}{23}\)