Given A, B, and C matrices, evaluate A-B+C.

Question 1(a)

Step 1: Separate the compound inequality into two inequalities.
The given inequality is $-4 < 2x - \frac{1}{3}(1 - \frac{1}{2}x) \le 1$.
This can be split into:
a) $-4 < 2x - \frac{1}{3}(1 - \frac{1}{2}x)$
b) $2x - \frac{1}{3}(1 - \frac{1}{2}x) \le 1$

Step 2: Solve the first inequality a).
$-4 < 2x - \frac{1}{3} + \frac{1}{6}x$
$-4 < \left(2 + \frac{1}{6}\right)x - \frac{1}{3}$
$-4 < \frac{13}{6}x - \frac{1}{3}$
Add $\frac{1}{3}$ to both sides:
$-4 + \frac{1}{3} < \frac{13}{6}x$
$-\frac{12}{3} + \frac{1}{3} < \frac{13}{6}x$
$-\frac{11}{3} < \frac{13}{6}x$
Multiply by $\frac{6}{13}$:
$-\frac{11}{3} \cdot \frac{6}{13} < x$
$-\frac{22}{13} < x$

Step 3: Solve the second inequality b).
$2x - \frac{1}{3}(1 - \frac{1}{2}x) \le 1$
$2x - \frac{1}{3} + \frac{1}{6}x \le 1$
$\left(2 + \frac{1}{6}\right)x - \frac{1}{3} \le 1$
$\frac{13}{6}x - \frac{1}{3} \le 1$
Add $\frac{1}{3}$ to both sides:
$\frac{13}{6}x \le 1 + \frac{1}{3}$
$\frac{13}{6}x \le \frac{3}{3} + \frac{1}{3}$
$\frac{13}{6}x \le \frac{4}{3}$
Multiply by $\frac{6}{13}$:
$x \le \frac{4}{3} \cdot \frac{6}{13}$
$x \le \frac{8}{13}$

Step 4: Combine the solutions from Step 2 and Step 3.
We have $x > -\frac{22}{13}$ and $x \le \frac{8}{13}$.
The solution set is $-\frac{22}{13} < x \le \frac{8}{13}$.

Step 5: Illustrate the solution on a number line.
The interval is $(-\frac{22}{13}, \frac{8}{13}]$.
On the number line, place an open circle at $-\frac{22}{13}$ (approximately $-1.69$) and a closed circle at $\frac{8}{13}$ (approximately $0.62$). Shade the region between these two points.

<--------------------------------------------------------------------->
      -2          -22/13          -1           0          8/13         1
<-----o----------------------------------------------------•------------>

-\frac{22{13} < x \le \frac{8}{13}}