Certainly! Let's solve these questions one by one.
### Question 10
We are given the equation [tex]\( |3 - 2x| = |5x + 2| + 1 \)[/tex] and asked to find the solution set.
From our solution process, we find that the equation has two distinct solutions:
- [tex]\( x = -\frac{4}{3} \)[/tex]
- [tex]\( x = 0 \)[/tex]
Hence, we can compare these solutions with each of the given choices:
A. [tex]\(\left\{-2, -\frac{4}{3}, 0, \frac{2}{7}\right\}\)[/tex] - This set does not match because it contains extraneous elements [tex]\(-2\)[/tex] and [tex]\(\frac{2}{7}\)[/tex].
B. [tex]\(\left\{-\frac{4}{3}, 0\right\}\)[/tex] - This is exactly the set of solutions we found.
C. [tex]\(\left\{-\frac{4}{3}, 0, \frac{2}{7}\right\}\)[/tex] - This set has an extraneous element [tex]\(\frac{2}{7}\)[/tex].
D. [tex]\(\{0\}\)[/tex] - This set only contains one of our solutions.
Thus, the correct answer is:
B. [tex]\(\left\{-\frac{4}{3}, 0\right\}\)[/tex]
### Question 11
We need to solve the inequality [tex]\(\left|\frac{2}{3} - 2x\right| \geq 2\)[/tex].
To handle this absolute value inequality, we break it into two separate inequalities based on the definition of absolute value:
1. [tex]\(\frac{2}{3} - 2x \geq 2\)[/tex]:
[tex]\[
\frac{2}{3} - 2x \geq 2 \implies -2x \geq 2 - \frac{2}{3} \implies -2x \geq \frac{6}{3} - \frac{2}{3} \implies -2x \geq \frac{4}{3}
\][/tex]
[tex]\[
x \leq -\frac{4}{6} \implies x \leq -\frac{2}{3}
\][/tex]
2. [tex]\(\frac{2}{3} - 2x \leq -2\)[/tex]:
[tex]\[
\frac{2}{3} - 2x \leq -2 \implies -2x \leq -2 - \frac{2}{3} \implies -2x \leq -\frac{6}{3} - \frac{2}{3} \implies -2x \leq -\frac{8}{3}
\][/tex]
[tex]\[
x \geq \frac{8}{6} \implies x \geq \frac{4}{3}
\][/tex]
Combining these two inequalities, we get:
[tex]\[
x \leq -\frac{2}{3} \text{ or } x \geq \frac{4}{3}
\][/tex]
Let's compare these intervals against each of the given choices:
A. [tex]\(\left(-\infty, \frac{1}{6}\right) \cup \left(\frac{13}{6}, \infty\right)\)[/tex] - This does not match our intervals.
B. [tex]\(\left(-\infty, \frac{1}{6}\right) \cup \left[\frac{13}{6}, \infty\right)\)[/tex] - This does not match either.
C. [tex]\(\left(-\infty, \frac{1}{6}\right] \cup \left(\frac{13}{6}, \infty\right)\)[/tex] - This does not include the boundary points we have.
D. [tex]\(\left(-\infty, \frac{1}{6}\right] \cup \left[\frac{13}{6}, \infty\right)\)[/tex] - This does not match because of the boundaries provided.
So, correcting for the intervals, the closest correct intervals are:
C. [tex]\(\left(-\infty, \frac{1}{6}\right] \cup \left(\frac{13}{6}, \infty\right)\)[/tex]
Our correction:
Answer should be:
[tex]\(\left(-\infty, -\frac{2}{3}\right] \cup \left[\frac{4}{3}, \infty\right)\)[/tex] which should match the intervals in choices correctly
