A certain grocery store sells oranges for $1.50 per pound and grapefruit for $1.20 per pound. You purchased a mix of oranges and grapefruits weighing 8 pounds total, which cost $10.35. If x is the number of pounds of grapefruit purchase, and y is the number of pounds of oranges purchased, the correct system of equations would be?

To solve this problem, you need to set up two equations based on the information provided:

1. The total weight of oranges and grapefruits is 8 pounds.
   \[
   x + y = 8
   \]
   Here, $x$ represents the number of pounds of grapefruit, and $y$ represents the number of pounds of oranges.

2. The total cost of the oranges and grapefruits is $10.35. The cost per pound for grapefruit is $1.20, and for oranges, it is $1.50. Therefore, the cost equation would be:
   \[
   1.20x + 1.50y = 10.35
   \]

So, the system of equations is:
\[
\begin{cases}
x + y = 8 \\
1.20x + 1.50y = 10.35
\end{cases}
\]

Would you like help solving this system of equations or more details on how to approach it?

### Expanding on the problem, here are 5 related questions:
1. What is the value of $x$ (the pounds of grapefruit) and $y$ (the pounds of oranges)?
2. If the cost per pound of oranges increased to $1.75, how would that change the system of equations?
3. What would the system of equations be if the total weight was 10 pounds instead of 8?
4. How much would 5 pounds of each fruit cost at the given prices?
5. What if you had purchased 10.35 pounds of fruit but still paid $10.35?

### Tip:
To solve systems of equations, substitution and elimination methods are two powerful techniques that can simplify complex problems like this one!

Learn how to solve a system of linear equations where you purchase 8 pounds of a mix of oranges and grapefruits for $10.35. This problem explores concepts in algebra, including creating equations based on weight and cost, and solving them using substitution or elimination methods. Ideal for Grades 8-10 students.

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