1. **Problem statement:** Solve the system of equations using the addition (elimination) method. 2. **Method:** The addition method involves adding or subtracting the equations to eliminate one variable, making it easier to solve for the other. 3. **System a:** $$\begin{cases} 3x + 2y = 6 \\ 2x - 2y = 14 \end{cases}$$ Add the two equations to eliminate $y$: $$ (3x + 2y) + (2x - 2y) = 6 + 14 $$ $$ 3x + 2x + \cancel{2y} - \cancel{2y} = 20 $$ $$ 5x = 20 $$ Divide both sides by 5: $$ x = \frac{20}{5} $$ $$ x = 4 $$ Substitute $x=4$ into the first equation: $$ 3(4) + 2y = 6 $$ $$ 12 + 2y = 6 $$ Subtract 12 from both sides: $$ 2y = 6 - 12 $$ $$ 2y = -6 $$ Divide both sides by 2: $$ y = \frac{-6}{2} $$ $$ y = -3 $$ **Final solution for a:** $x=4$, $y=-3$. 4. **System b:** $$\begin{cases} 0.3x + 0.5y = 3 \\ 1.5x - 2y = 1.5 \end{cases}$$ Multiply the first equation by 4 to align $y$ coefficients: $$ 4(0.3x + 0.5y) = 4(3) $$ $$ 1.2x + 2y = 12 $$ Add this to the second equation: $$ (1.2x + 2y) + (1.5x - 2y) = 12 + 1.5 $$ $$ 1.2x + 1.5x + \cancel{2y} - \cancel{2y} = 13.5 $$ $$ 2.7x = 13.5 $$ Divide both sides by 2.7: $$ x = \frac{13.5}{2.7} $$ $$ x = 5 $$ Substitute $x=5$ into the first original equation: $$ 0.3(5) + 0.5y = 3 $$ $$ 1.5 + 0.5y = 3 $$ Subtract 1.5: $$ 0.5y = 1.5 $$ Divide by 0.5: $$ y = \frac{1.5}{0.5} $$ $$ y = 3 $$ **Final solution for b:** $x=5$, $y=3$. 5. **System c:** $$\begin{cases} -2x + 6y = 8 \\ 3x - 9y = -12 \end{cases}$$ Multiply the first equation by 3 and the second by 2 to align $x$ coefficients: $$ 3(-2x + 6y) = 3(8) \Rightarrow -6x + 18y = 24 $$ $$ 2(3x - 9y) = 2(-12) \Rightarrow 6x - 18y = -24 $$ Add the two equations: $$ (-6x + 18y) + (6x - 18y) = 24 + (-24) $$ $$ \cancel{-6x} + \cancel{6x} + \cancel{18y} - \cancel{18y} = 0 $$ $$ 0 = 0 $$ This means the two equations are dependent and represent the same line. Express $x$ in terms of $y$ from the first equation: $$ -2x + 6y = 8 $$ $$ -2x = 8 - 6y $$ $$ x = \frac{6y - 8}{2} = 3y - 4 $$ **Final solution for c:** Infinite solutions along the line $x = 3y - 4$. 6. **System d:** $$\begin{cases} 3x - 7y = -14 \\ 5x + 2y = 45 \end{cases}$$ Multiply the first equation by 2 and the second by 7 to align $y$ coefficients: $$ 2(3x - 7y) = 2(-14) \Rightarrow 6x - 14y = -28 $$ $$ 7(5x + 2y) = 7(45) \Rightarrow 35x + 14y = 315 $$ Add the two equations: $$ (6x - 14y) + (35x + 14y) = -28 + 315 $$ $$ 6x + 35x + \cancel{-14y} + \cancel{14y} = 287 $$ $$ 41x = 287 $$ Divide both sides by 41: $$ x = \frac{287}{41} $$ $$ x = 7 $$ Substitute $x=7$ into the first original equation: $$ 3(7) - 7y = -14 $$ $$ 21 - 7y = -14 $$ Subtract 21: $$ -7y = -35 $$ Divide by -7: $$ y = 5 $$ **Final solution for d:** $x=7$, $y=5$. 7. **System e:** $$\begin{cases} x - 0.5y = -3 \\ 2x - y = 6 \end{cases}$$ Multiply the first equation by 2: $$ 2(x - 0.5y) = 2(-3) $$ $$ 2x - y = -6 $$ Subtract this from the second equation: $$ (2x - y) - (2x - y) = 6 - (-6) $$ $$ \cancel{2x} - \cancel{2x} - y + y = 12 $$ $$ 0 = 12 $$ This is a contradiction, so no solution. **Final solution for e:** No solution. 8. **System f:** $$\begin{cases} 2x + 3y = 3 \\ \frac{2}{3}x + y = 1 \end{cases}$$ Multiply the second equation by 3: $$ 3 \left( \frac{2}{3}x + y \right) = 3(1) $$ $$ 2x + 3y = 3 $$ This is the same as the first equation, so infinite solutions along the line $2x + 3y = 3$. **Final solution for f:** Infinite solutions along $2x + 3y = 3$. **Summary:** - a) $x=4$, $y=-3$ - b) $x=5$, $y=3$ - c) Infinite solutions: $x=3y-4$ - d) $x=7$, $y=5$ - e) No solution - f) Infinite solutions: $2x + 3y = 3$