1. **Problem Statement:** Solve the following systems of linear equations using the Addition (Elimination) method. --- ### a. \begin{cases} 2x - 3y = 5 \\ (1x - 2)y = 6 \end{cases} Note: The second equation seems ambiguous as written "(1x - 2)y = 6". Assuming it means $x - 2y = 6$. **Step 1:** Write the system clearly: $$\begin{cases} 2x - 3y = 5 \\ x - 2y = 6 \end{cases}$$ **Step 2:** Multiply the second equation by $-2$ to align coefficients of $x$ for elimination: $$\begin{cases} 2x - 3y = 5 \\ -2x + 4y = -12 \end{cases}$$ **Step 3:** Add the two equations: $$ (2x - 3y) + (-2x + 4y) = 5 + (-12) $$ $$ \cancel{2x} - 3y - \cancel{2x} + 4y = -7 $$ $$ y = -7 $$ **Step 4:** Substitute $y = -7$ into the second original equation: $$ x - 2(-7) = 6 $$ $$ x + 14 = 6 $$ $$ x = 6 - 14 = -8 $$ **Solution for a:** $x = -8$, $y = -7$ --- ### b. $$\begin{cases} 2x - 7y = 2 \\ 3x + y = -20 \end{cases}$$ **Step 1:** Multiply the second equation by 7 to align $y$ coefficients: $$\begin{cases} 2x - 7y = 2 \\ 21x + 7y = -140 \end{cases}$$ **Step 2:** Add the equations: $$ (2x - 7y) + (21x + 7y) = 2 + (-140) $$ $$ 2x + 21x - 7y + 7y = -138 $$ $$ 23x = -138 $$ $$ x = \frac{-138}{23} = -6 $$ **Step 3:** Substitute $x = -6$ into the second original equation: $$ 3(-6) + y = -20 $$ $$ -18 + y = -20 $$ $$ y = -20 + 18 = -2 $$ **Solution for b:** $x = -6$, $y = -2$ --- ### c. $$\begin{cases} 2x + 3y = 5 \\ 3x + 5y = 7 \end{cases}$$ **Step 1:** Multiply the first equation by 3 and the second by -2 to align $x$ coefficients: $$\begin{cases} 6x + 9y = 15 \\ -6x - 10y = -14 \end{cases}$$ **Step 2:** Add the equations: $$ (6x + 9y) + (-6x - 10y) = 15 + (-14) $$ $$ \cancel{6x} + 9y - \cancel{6x} - 10y = 1 $$ $$ -y = 1 $$ $$ y = -1 $$ **Step 3:** Substitute $y = -1$ into the first original equation: $$ 2x + 3(-1) = 5 $$ $$ 2x - 3 = 5 $$ $$ 2x = 8 $$ $$ x = 4 $$ **Solution for c:** $x = 4$, $y = -1$ --- ### d. $$\begin{cases} 9x + 3y = 38 \\ 5x + y = 22 \end{cases}$$ **Step 1:** Multiply the second equation by -3 to align $y$ coefficients: $$\begin{cases} 9x + 3y = 38 \\ -15x - 3y = -66 \end{cases}$$ **Step 2:** Add the equations: $$ (9x + 3y) + (-15x - 3y) = 38 + (-66) $$ $$ 9x - 15x + 3y - 3y = -28 $$ $$ -6x = -28 $$ $$ x = \frac{-28}{-6} = \frac{14}{3} $$ **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 for d:** $x = \frac{14}{3}$, $y = -\frac{4}{3}$