1. **Problem 12:** Identify the linear system modeled by the balance scales. Given: Each small square represents 2 kg. - Left scale: Left side has 1 block of $x$. - Right side has 1 block of $x$, 1 block of $y$, and 5 small squares. Weight on right side = $x + y + 5 \times 2 = x + y + 10$. Since scales balance, left side weight = right side weight: $$x = x + y + 10$$ Simplify: $$x - x = y + 10$$ $$0 = y + 10$$ $$y = -10$$ This seems contradictory, so let's re-express carefully. Actually, the problem states the left side has 1 block $x$, right side has 1 block $x$, 1 block $y$, and 5 small squares (each 2 kg), so total right side weight: $$x + y + 5 \times 2 = x + y + 10$$ Balance means: $$x = x + y + 10$$ Subtract $x$ from both sides: $$\cancel{x} = \cancel{x} + y + 10$$ $$0 = y + 10$$ $$y = -10$$ This negative weight is not possible, so likely the problem means the left side has 2 blocks of $x$ (since the options have $2x$), or the problem is about the equations given. Check the options: - a. $2x + y = 14$ and $x + 3y = 12$ - b. $2x + y = 7$ and $x + 3y = 6$ - c. $x + 2y = 14$ and $3x + y = 12$ - d. $2x + y = 6$ and $x + 3y = 7$ From the description: - Left scale: Left side has 1 block $x$. - Right side has 1 block $x$, 1 block $y$, and 5 small squares (5*2=10 kg). So total right side weight: $x + y + 10$. Balance means: $$x = x + y + 10$$ $$0 = y + 10$$ $$y = -10$$ Negative weight is impossible, so maybe the left side has 2 blocks $x$ (2x), right side has $y$ and 7 small squares (7*2=14 kg) for the first equation. Similarly, for the right scale: - Left side has 1 block $y$. - Right side has 1 block $x$, 1 block $y$, and 6 small squares (6*2=12 kg). Balance means: $$y = x + y + 12$$ $$0 = x + 12$$ $$x = -12$$ Again negative weight. Looking at the options, option a matches the weights: - $2x + y = 14$ - $x + 3y = 12$ So the system modeled is option a. --- 2. **Problem 13:** Find $y$ given solution $(-28, y)$ for system: $$\frac{1}{2}x - \frac{1}{4}y = -20$$ $$\frac{8}{9}x - 4y = -\frac{1088}{9}$$ Step 1: Substitute $x = -28$ into first equation: $$\frac{1}{2}(-28) - \frac{1}{4}y = -20$$ $$-14 - \frac{1}{4}y = -20$$ Step 2: Isolate $y$: $$- \frac{1}{4}y = -20 + 14 = -6$$ $$y = \frac{-6}{-\frac{1}{4}} = -6 \times -4 = 24$$ Step 3: Verify with second equation: $$\frac{8}{9}(-28) - 4y = -\frac{1088}{9}$$ $$-\frac{224}{9} - 4y = -\frac{1088}{9}$$ Add $\frac{224}{9}$ to both sides: $$-4y = -\frac{1088}{9} + \frac{224}{9} = -\frac{864}{9} = -96$$ Divide both sides by $-4$: $$y = \frac{-96}{-4} = 24$$ Answer: $y = 24$ (option b). --- 3. **Problem 14:** Write equivalent system with integer coefficients for: $$\frac{5}{8}x + \frac{3}{4}y = \frac{35}{4}$$ $$\frac{8}{9}x + 5y = \frac{643}{9}$$ Step 1: Multiply first equation by 8 to clear denominators: $$8 \times \left( \frac{5}{8}x + \frac{3}{4}y \right) = 8 \times \frac{35}{4}$$ $$5x + 6y = 70$$ Step 2: Multiply second equation by 9: $$9 \times \left( \frac{8}{9}x + 5y \right) = 9 \times \frac{643}{9}$$ $$8x + 45y = 643$$ Answer: option b. --- 4. **Problem 15:** Determine number of solutions for system: $$5x - 7y = 41$$ $$-15x + 21y = 3$$ Step 1: Multiply first equation by 3: $$3(5x - 7y) = 3(41)$$ $$15x - 21y = 123$$ Step 2: Compare with second equation: $$-15x + 21y = 3$$ Add both equations: $$(15x - 21y) + (-15x + 21y) = 123 + 3$$ $$0 = 126$$ Contradiction means no solution. Answer: option c. --- **Final answers:** 12: a 13: b 14: b 15: c