1. **Problem Statement:** Solve the following systems of linear equations using the Addition (Elimination) method. --- ### a. \begin{cases} 2x - 3y = 5 \\ 1x - 2y = 6 \end{cases} **Step 1:** Multiply the second equation by 2 to align coefficients of $x$: $$\begin{cases} 2x - 3y = 5 \\ 2x - 4y = 12 \end{cases}$$ **Step 2:** Subtract the first equation from the second: $$\cancel{2x} - 4y - (\cancel{2x} - 3y) = 12 - 5$$ $$-4y + 3y = 7$$ $$-y = 7 \Rightarrow y = -7$$ **Step 3:** Substitute $y = -7$ into the second original equation: $$x - 2(-7) = 6$$ $$x + 14 = 6$$ $$x = 6 - 14 = -8$$ **Solution:** $x = -8$, $y = -7$ --- ### b. \begin{cases} 2x - 7y = 2 \\ 3x + y = -20 \end{cases} **Step 1:** Multiply the second equation by 7: $$\begin{cases} 2x - 7y = 2 \\ 21x + 7y = -140 \end{cases}$$ **Step 2:** Add the two equations: $$(2x + 21x) + (-7y + 7y) = 2 + (-140)$$ $$23x = -138$$ $$x = \frac{-138}{23} = -6$$ **Step 3:** Substitute $x = -6$ into the second original equation: $$3(-6) + y = -20$$ $$-18 + y = -20$$ $$y = -20 + 18 = -2$$ **Solution:** $x = -6$, $y = -2$ --- ### c. \begin{cases} 2x + 3y = 5 \\ 3x + 5y = 7 \end{cases} **Step 1:** Multiply the first equation by 3 and the second by 2: $$\begin{cases} 6x + 9y = 15 \\ 6x + 10y = 14 \end{cases}$$ **Step 2:** Subtract the first from the second: $$\cancel{6x} + 10y - (\cancel{6x} + 9y) = 14 - 15$$ $$10y - 9y = -1$$ $$y = -1$$ **Step 3:** Substitute $y = -1$ into the first original equation: $$2x + 3(-1) = 5$$ $$2x - 3 = 5$$ $$2x = 8$$ $$x = 4$$ **Solution:** $x = 4$, $y = -1$ --- ### d. \begin{cases} 9x + 3y = 38 \\ 5x + y = 22 \end{cases} **Step 1:** Multiply the second equation by 3: $$\begin{cases} 9x + 3y = 38 \\ 15x + 3y = 66 \end{cases}$$ **Step 2:** Subtract the first from the second: $$15x + 3y - (9x + 3y) = 66 - 38$$ $$(15x - 9x) + (3y - 3y) = 28$$ $$6x = 28$$ $$x = \frac{28}{6} = \frac{14}{3}$$ **Step 3:** Substitute $x = \frac{14}{3}$ into the second original equation: $$5 \times \frac{14}{3} + y = 22$$ $$\frac{70}{3} + y = 22$$ $$y = 22 - \frac{70}{3} = \frac{66}{3} - \frac{70}{3} = -\frac{4}{3}$$ **Solution:** $x = \frac{14}{3}$, $y = -\frac{4}{3}$ --- ### e. \begin{cases} -3x + y = -9 \\ -7x + 4y = 6 \end{cases} **Step 1:** Multiply the first equation by 4: $$\begin{cases} -12x + 4y = -36 \\ -7x + 4y = 6 \end{cases}$$ **Step 2:** Subtract the second from the first: $$(-12x + 4y) - (-7x + 4y) = -36 - 6$$ $$-12x + 4y + 7x - 4y = -42$$ $$-5x = -42$$ $$x = \frac{-42}{-5} = \frac{42}{5}$$ **Step 3:** Substitute $x = \frac{42}{5}$ into the first original equation: $$-3 \times \frac{42}{5} + y = -9$$ $$-\frac{126}{5} + y = -9$$ $$y = -9 + \frac{126}{5} = -\frac{45}{5} + \frac{126}{5} = \frac{81}{5}$$ **Solution:** $x = \frac{42}{5}$, $y = \frac{81}{5}$