To solve this problem, let's define the variables:

• Let $r$ represent the cost of one packet of rice in dollars.
• Let $c$ represent the cost of one cake in dollars.

We are given two situations that can be translated into algebraic equations:

1. First situation: Kim bought 3 packets of rice and 6 cakes for $33. $3r + 6c = 33$

2. Second situation: Kim bought 2 packets of rice and 6 cakes for $30. $2r + 6c = 30$

Now we have a system of two linear equations:

3r + 6c &= 33 \quad \text{(Equation 1)} \\ 2r + 6c &= 30 \quad \text{(Equation 2)} \end{align*}$$ ### Step 1: Subtract Equation 2 from Equation 1 By subtracting the second equation from the first, we can eliminate $$c$$ and solve for $$r$$: $$(3r + 6c) - (2r + 6c) = 33 - 30$$ Simplify the equation: $$3r - 2r + 6c - 6c = 3$$ $$r = 3$$ So, the cost of one packet of rice, $$r$$, is **$3**. ### Step 2: Substitute the value of $$r$$ into one of the equations Let's substitute $$r = 3$$ into Equation 2 to solve for $$c$$: $$2r + 6c = 30$$ Substitute $$r = 3$$: $$2(3) + 6c = 30$$ Simplify: $$6 + 6c = 30$$ Subtract 6 from both sides: $$6c = 24$$ Divide by 6: $$c = 4$$ So, the cost of one cake, $$c$$, is **$4**. ### Summary - **Cost of one packet of rice**: $3 - **Cost of one cake**: $4 Would you like further details on these steps, or do you have any more questions? Here are some related questions you might find interesting: 1. If Kim bought 5 packets of rice and 10 cakes, how much would it cost? 2. How much would 7 packets of rice and 2 cakes cost? 3. If the price of cakes increased by 20%, how would that affect the total cost in each scenario? 4. If Kim only bought cakes on the third day and spent $24, how many cakes did she buy? 5. What is the difference in cost if Kim buys 8 packets of rice instead of 4? **Tip:** When dealing with systems of equations, always look for ways to eliminate one variable by adding or subtracting equations. This simplifies finding the value of the remaining variables!