The problem in the image asks to solve simultaneous equations using the elimination method. I'll focus on part (c) first and proceed step by step to solve it.

(c) Solve the simultaneous equations:

$\frac{x}{3} + \frac{y}{2} = 3 \quad \text{(1)}$ $\frac{x}{3} - \frac{y}{5} = 12 \quad \text{(2)}$

Step 1: Eliminate the fractions.

To eliminate the fractions, we can multiply both equations by the least common denominator (LCD) of the denominators in each equation.

• For equation (1), the LCD of 3 and 2 is 6, so multiply everything by 6: $6 \left(\frac{x}{3} + \frac{y}{2} \right) = 6 \cdot 3$ This simplifies to: $2x + 3y = 18 \quad \text{(3)}$

• For equation (2), the LCD of 3 and 5 is 15, so multiply everything by 15: $15 \left(\frac{x}{3} - \frac{y}{5} \right) = 15 \cdot 12$ This simplifies to: $5x - 3y = 180 \quad \text{(4)}$

Step 2: Eliminate one variable.

Now we can eliminate either $x$ or $y$. Adding equations (3) and (4) will eliminate $y$ because the coefficients of $y$ are opposites:

• Add equations (3) and (4): $(2x + 3y) + (5x - 3y) = 18 + 180$ Simplifying: $7x = 198$ Solving for $x$: $x = \frac{198}{7} = 28.29$

Step 3: Solve for $y$.

Substitute $x = 28.29$ into equation (3) to solve for $y$: $2(28.29) + 3y = 18$ This simplifies to: $56.58 + 3y = 18$ $3y = 18 - 56.58$ $3y = -38.58$ $y = \frac{-38.58}{3} \approx -12.86$

Thus, the solution is approximately $x \approx 28.29$ and $y \approx -12.86$.

Do you want me to proceed with solving part (d) as well? Let me know if you need more details or have questions.

Here are 5 related questions for practice:

1. What are the steps of the elimination method for solving simultaneous equations?
2. How do you handle fractions in simultaneous equations?
3. Can simultaneous equations always be solved using the elimination method?
4. Why do we multiply equations by a common denominator in fraction-based equations?
5. How can we verify the solution to a system of equations?

Tip: Always simplify the system by eliminating fractions or decimals first for easier calculation.