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This problem can be solved using a system of linear equations. Let's define the following variables:

- Let $ x $ be the cost per pound of rib steak.
- Let $ y $ be the cost per pound of hamburger meat.

From the information in the problem, we have two pieces of information that can be turned into two equations:

1. $ 2x + 6y = 12.30 $ (since 2 pounds of rib steak and 6 pounds of hamburger meat cost $12.30)
2. $ 3x + 2y = 9.70 $ (since 3 pounds of rib steak and 2 pounds of hamburger meat cost $9.70)

We now have the system of equations:

\[
\begin{aligned}
2x + 6y &= 12.30 \\
3x + 2y &= 9.70
\end{aligned}
\]

We can solve this system using either substitution or elimination. Let's use the elimination method by multiplying the second equation by 3 and the first equation by 1, so that the coefficients of $ y $ will be the same:

\[
\begin{aligned}
2x + 6y &= 12.30 \quad \text{(equation 1)} \\
9x + 6y &= 29.10 \quad \text{(equation 2 multiplied by 3)}
\end{aligned}
\]

Now, subtract equation 1 from this new equation 2:

\[
(9x + 6y) - (2x + 6y) = 29.10 - 12.30
\]

Simplifying:

\[
7x = 16.80
\]

Solving for $ x $:

\[
x = \frac{16.80}{7} = 2.40
\]

Now that we know the cost per pound of rib steak is $2.40, we substitute this value into one of the original equations to find $ y $. Let's use the second equation:

\[
3x + 2y = 9.70
\]

Substitute $ x = 2.40 $:

\[
3(2.40) + 2y = 9.70
\]

\[
7.20 + 2y = 9.70
\]

Subtract 7.20 from both sides:

\[
2y = 2.50
\]

Solving for $ y $:

\[
y = \frac{2.50}{2} = 1.25
\]

Thus, the cost per pound of rib steak is **$2.40**, and the cost per pound of hamburger meat is **$1.25**.

Would you like further details or have any questions?

Here are 5 related questions to think about:
1. How would the solution change if the prices were increased by 10%?
2. Can you solve the system using substitution instead of elimination?
3. What if the ratio of rib steak to hamburger meat was different? How would the equations change?
4. How would you set up a similar system if you knew only the total weight and total cost, but not the individual weights?
5. Could this system be solved graphically? What would the graph look like?

**Tip:** When solving systems of equations, elimination is often faster for systems with similar coefficients, while substitution works well when one equation is easily solved for a variable.

If 2 pounds of rib steak and 6 pounds of hamburger meat costs $12.30 and 3 pounds of rib steak and 2 pounds of hamburger meat costs $9.70, what is the cost per pound of each type of meat?

Learn how to find the cost per pound of rib steak and hamburger meat by solving a system of linear equations. This problem demonstrates algebraic techniques for solving real-life situations involving linear equations. Suitable for students in Grades 7-10.

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