Solve the following systems of simultaneous equations.

Here are the solutions to the problems:

1. Solve the system of the simultaneous equations

• i. Given the system:

  $$\begin{cases} 3x - y = 1 \quad &(1) \\ x - y = -3 \quad &(2) \end{cases}$$

  Step 1: Subtract equation (2) from equation (1) to eliminate $y$.

  $$(3x - y) - (x - y) = 1 - (-3)$$ $$3x - y - x + y = 1 + 3$$ $$2x = 4$$

  Step 2: Solve for $x$.

  $$x = \frac{4}{2}$$ $$x = 2$$

  Step 3: Substitute $x=2$ into equation (2) to find $y$.

  $$2 - y = -3$$ $$-y = -3 - 2$$ $$-y = -5$$ $$y = 5$$

  The solution is $\boxed{x=2, y=5}$.

• ii. Given the system:

  $$\begin{cases} 3x - 4y = 0 \quad &(1) \\ x - y = 2 \quad &(2) \end{cases}$$

  Step 1: From equation (2), express $x$ in terms of $y$.

  $$x = y + 2$$

  Step 2: Substitute this expression for $x$ into equation (1).

  $$3(y+2) - 4y = 0$$ $$3y + 6 - 4y = 0$$ $$-y + 6 = 0$$ $$-y = -6$$ $$y = 6$$

  Step 3: Substitute $y=6$ back into the expression for $x$.

  $$x = 6 + 2$$ $$x = 8$$

  The solution is $\boxed{x=8, y=6}$.

• iii. Given the system:

  $$\begin{cases} x = 4 - \frac{3}{2}y \quad &(1) \\ -3x + y = -1 \quad &(2) \end{cases}$$

  Step 1: Substitute the expression for $x$ from equation (1) into equation (2).

  $$-3\left(4 - \frac{3}{2}y\right) + y = -1$$ $$-12 + \frac{9}{2}y + y = -1$$

  Step 2: Combine the $y$ terms.

  $$-12 + \frac{9}{2}y + \frac{2}{2}y = -1$$ $$-12 + \frac{11}{2}y = -1$$

  Step 3: Add 12 to both sides.

  $$\frac{11}{2}y = -1 + 12$$ $$\frac{11}{2}y = 11$$

  Step 4: Multiply by $\frac{2}{11}$ to solve for $y$.

  $$y = 11 \times \frac{2}{11}$$ $$y = 2$$

  Step 5: Substitute $y=2$ back into equation (1) to find $x$.

  $$x = 4 - \frac{3}{2}(2)$$ $$x = 4 - 3$$ $$x = 1$$

  The solution is $\boxed{x=1, y=2}$.

2. Solve the inequalities

• i.

  $$\frac{1}{3}x - \frac{1}{2}(x-3) < 1$$

  Step 1: Distribute $\frac{1}{2}$.

  $$\frac{1}{3}x - \frac{1}{2}x + \frac{3}{2} < 1$$

  Step 2: Find a common denominator for the $x$ terms (LCM of 3 and 2 is 6).

  $$\frac{2}{6}x - \frac{3}{6}x + \frac{3}{2} < 1$$ $$-\frac{1}{6}x + \frac{3}{2} < 1$$

  Step 3: Subtract $\frac{3}{2}$ from both sides.

  $$-\frac{1}{6}x < 1 - \frac{3}{2}$$ $$-\frac{1}{6}x < \frac{2}{2} - \frac{3}{2}$$ $$-\frac{1}{6}x < -\frac{1}{2}$$

  Step 4: Multiply both sides by -6 and reverse the inequality sign.

  $$x > \left(-\frac{1}{2}\right) \times (-6)$$ $$x > 3$$

  The solution is $\boxed{x > 3}$.

• ii.

  $$\frac{x+4}{2x+2} \le 1$$

  Step 1: Subtract 1 from both sides.

  $$\frac{x+4}{2x+2} - 1 \le 0$$

  Step 2: Find a common denominator.

  $$\frac{x+4}{2x+2} - \frac{2x+2}{2x+2} \le 0$$ $$\frac{x+4 - (2x+2)}{2x+2} \le 0$$ $$\frac{x+4 - 2x - 2}{2x+2} \le 0$$ $$\frac{-x+2}{2x+2} \le 0$$

  Step 3: Find critical points by setting the numerator and denominator to zero. Numerator: $-x+2 = 0 \implies x = 2$ Denominator: $2x+2 = 0 \implies 2x = -2 \implies x = -1$

  Step 4: Test intervals $(-\infty, -1)$, $(-1, 2]$, and $[2, \infty)$.

  • For $x < -1$ (e.g., $x=-2$): $\frac{-(-2)+2}{2(-2)+2} = \frac{4}{-2} = -2 \le 0$. This interval is part of the solution.
  • For $-1 < x \le 2$ (e.g., $x=0$): $\frac{-0+2}{2(0)+2} = \frac{2}{2} = 1 \not\le 0$. This interval is not part of the solution.
  • For $x > 2$ (e.g., $x=3$): $\frac{-3+2}{2(3)+2} = \frac{-1}{8} \le 0$. This interval is part of the solution. Note that $x=-1$ is excluded because it makes the denominator zero. $x=2$ is included because the inequality is $\le$. The solution is $\boxed{x < -1 \text{ or } x \ge 2}$.

• iii.

  $$\frac{4}{5}x \le \frac{7}{5} + \frac{2}{5}x$$

  Step 1: Subtract $\frac{2}{5}x$ from both sides.

  $$\frac{4}{5}x - \frac{2}{5}x \le \frac{7}{5}$$ $$\frac{2}{5}x \le \frac{7}{5}$$

  Step 2: Multiply both sides by $\frac{5}{2}$.

  $$x \le \frac{7}{5} \times \frac{5}{2}$$ $$x \le \frac{7}{2}$$

  The solution is $\boxed{x \le \frac{7}{2}}$.

• iv.

  $$\frac{1}{2}\left(5x - 4 \ge x - \frac{11}{2}\right)$$

  Step 1: Multiply both sides by 2 to clear the fraction.

  $$5x - 4 \ge 2\left(x - \frac{11}{2}\right)$$ $$5x - 4 \ge 2x - 11$$

  Step 2: Subtract $2x$ from both sides.

  $$5x - 2x - 4 \ge -11$$ $$3x - 4 \ge -11$$

  Step 3: Add 4 to both sides.

  $$3x \ge -11 + 4$$ $$3x \ge -7$$

  Step 4: Divide by 3.

  $$x \ge -\frac{7}{3}$$

  The solution is $\boxed{x \ge -\frac{7}{3}}$.

• v.

  $$5x - 2 \le 3x + 6$$

  Step 1: Subtract $3x$ from both sides.

  $$5x - 3x - 2 \le 6$$ $$2x - 2 \le 6$$

  Step 2: Add 2 to both sides.

  $$2x \le 6 + 2$$ $$2x \le 8$$

  Step 3: Divide by 2.

  $$x \le 4$$

  The solution is $\boxed{x \le 4}$.

• vi.

  $$x - 7 \ge -5$$

  Step 1: Add 7 to both sides.

  $$x \ge -5 + 7$$ $$x \ge 2$$

  The solution is $\boxed{x \ge 2}$.

3. Solve for x

• i.

  $$\frac{3}{x-2} - \frac{4}{x+1} = 0$$

  Step 1: Add $\frac{4}{x+1}$ to both sides.

  $$\frac{3}{x-2} = \frac{4}{x+1}$$

  Step 2: Cross-multiply.

  $$3(x+1) = 4(x-2)$$ $$3x + 3 = 4x - 8$$

  Step 3: Subtract $3x$ from both sides.

  $$3 = 4x - 3x - 8$$ $$3 = x - 8$$

  Step 4: Add 8 to both sides.

  $$3 + 8 = x$$ $$x = 11$$

  Check for restrictions: $x \ne 2$ and $x \ne -1$. Our solution $x=11$ is valid. The solution is $\boxed{x=11}$.

• ii.

  $$\frac{3x-5}{9} + 2x = \frac{1}{3}$$

  Step 1: Find a common denominator for all terms (LCM of 9 and 3 is 9).

  $$\frac{3x-5}{9} + \frac{18x}{9} = \frac{3}{9}$$

  Step 2: Multiply the entire equation by 9.

  $$3x - 5 + 18x = 3$$

  Step 3: Combine like terms.

  $$21x - 5 = 3$$

  Step 4: Add 5 to both sides.

  $$21x = 3 + 5$$ $$21x = 8$$

  Step 5: Divide by 21.

  $$x = \frac{8}{21}$$

  The solution is $\boxed{x=\frac{8}{21}}$.

• iii.

  $$\frac{x}{7} - \frac{x}{5} = \frac{10}{7}$$

  Step 1: Find a common denominator for the $x$ terms (LCM of 7 and 5 is 35).

  $$\frac{5x}{35} - \frac{7x}{35} = \frac{10}{7}$$ $$$$