Solve the inequality below and write down the integral values that satisfy the equality

Here's the solution for questions 9 and 10:

9. Solve the inequality below and write down the integral values that satisfy the equality $-3x + 2 < x + 6 \le 17 - 2x$

Step 1: Split the compound inequality into two separate inequalities. $-3x + 2 < x + 6 \quad and \quad x + 6 \le 17 - 2x$

Step 2: Solve the first inequality. \begin{align*} -3x + 2 &< x + 6 \\ -3x - x &< 6 - 2 \\ -4x &< 4 \end{align*} Divide by $-4$ and reverse the inequality sign: $x > \frac{4}{-4} \\ x > -1$

Step 3: Solve the second inequality. \begin{align*} x + 6 &\le 17 - 2x \\ x + 2x &\le 17 - 6 \\ 3x &\le 11 \end{align*} Divide by $3$: $x \le \frac{11}{3}$

Step 4: Combine the solutions and identify the integral values. The combined inequality is $-1 < x \le \frac{11}{3}$. Since $\frac{11}{3} \approx 3.67$, the integral values (integers) that satisfy this condition are those greater than $-1$ and less than or equal to $3.67$. The integral values are $\boxed{0, 1, 2, 3}$.

10. Find the equation of a line which passes through the point $(2, 3)$ and is perpendicular to $y - 3x + 1 = 0$, giving your answer in the form $y = mx + c$.

Step 1: Find the slope of the given line. The given equation is $y - 3x + 1 = 0$. Rearrange it into the form $y = mx + c$: $y = 3x - 1$ The slope of this line, $m_1$, is $3$.

Step 2: Find the slope of the perpendicular line. For two lines to be perpendicular, the product of their slopes must be $-1$. Let $m_2$ be the slope of the required line. $m_1 \cdot m_2 = -1 \\ 3 \cdot m_2 = -1 \\ m_2 = -\frac{1}{3}$

Step 3: Use the point-slope form to find the equation of the required line. The line passes through $(x_1, y_1) = (2, 3)$ and has a slope $m_2 = -\frac{1}{3}$. The point-slope form is $y - y_1 = m(x - x_1)$. $y - 3 = -\frac{1}{3}(x - 2)$

Step 4: Convert the equation to the $y = mx + c$ form. $y - 3 = -\frac{1}{3}x + \frac{2}{3} \\ y = -\frac{1}{3}x + \frac{2}{3} + 3 \\ y = -\frac{1}{3}x + \frac{2}{3} + \frac{9}{3} \\ y = -\frac{1}{3}x + \frac{11}{3}$ The equation of the line is $\boxed{y = -\frac{1}{3}x + \frac{11}{3}}$.

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