Delaney would like to make a 5 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds. Let *p* represent the number of pounds of peanuts needed to make the new mixture, and let *m* represent the number of pounds of the 80% almond-20% peanut mixture.

We can solve this problem by setting up a system of equations to represent the mixture of peanuts and almonds in the final 5-pound nut mixture. Let's break it down step by step:

### Given:
- Delaney wants a total of **5 pounds** of the nut mixture.
- The final mixture should be **60% peanuts** and **40% almonds**.
- She has two ingredients:
  1. **Pure peanuts** (100% peanuts).
  2. A mixture that is **20% peanuts** and **80% almonds**.

Let:
- \( p \) = the number of pounds of pure peanuts.
- \( m \) = the number of pounds of the 20% peanut and 80% almond mixture.

### Total Weight Equation:
The total weight of the mixture must equal 5 pounds:
\[
p + m = 5
\]

### Peanut Content Equation:
The final mixture should be 60% peanuts. So, the amount of peanuts from the two sources (pure peanuts and the peanut-almond mixture) must add up to 60% of 5 pounds, which is \( 0.60 \times 5 = 3 \) pounds of peanuts.

- From the pure peanuts, we get \( p \) pounds of peanuts.
- From the 20% peanut-almond mixture, we get \( 0.20 \times m \) pounds of peanuts.

Thus, the equation for the total peanut content is:
\[
p + 0.20m = 3
\]

### System of Equations:
Now we have the following system of equations:
1. \( p + m = 5 \)
2. \( p + 0.20m = 3 \)

### Solving the System:

#### Step 1: Solve for \( p \) in the first equation:
From the first equation \( p + m = 5 \), we can solve for \( p \):
\[
p = 5 - m
\]

#### Step 2: Substitute \( p = 5 - m \) into the second equation:
Substitute this expression for \( p \) into the second equation \( p + 0.20m = 3 \):
\[
(5 - m) + 0.20m = 3
\]

#### Step 3: Simplify and solve for \( m \):
\[
5 - m + 0.20m = 3
\]
\[
5 - 0.80m = 3
\]
\[
-0.80m = 3 - 5
\]
\[
-0.80m = -2
\]
\[
m = \frac{-2}{-0.80} = 2.5
\]

#### Step 4: Solve for \( p \):
Now that we know \( m = 2.5 \), substitute this value into the equation \( p = 5 - m \):
\[
p = 5 - 2.5 = 2.5
\]

### Final Answer:
- Delaney needs **2.5 pounds of pure peanuts**.
- She also needs **2.5 pounds of the 20% peanut and 80% almond mixture**.

Would you like more details or have any questions?

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Here are 5 related questions to expand this information:
1. What if Delaney wanted the final mixture to be 70% peanuts instead of 60%?
2. How would the system of equations change if the peanut content in the mixture was 30% instead of 20%?
3. What is the minimum amount of peanuts Delaney can use if she only has the peanut-almond mixture?
4. How would the solution change if the final mixture needed to be 10 pounds instead of 5 pounds?
5. How many pounds of almonds are in the final mixture?

**Tip:** Solving systems of linear equations can often be simplified by substitution or elimination techniques—practice both methods to improve your problem-solving efficiency!

Delaney would like to make a 5 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds. Let p represent the number of pounds of peanuts needed to make the new mixture, and let m represent the number of pounds of the 80% almond-20% peanut mixture.

Learn how to solve a system of equations to determine the number of pounds of peanuts and a peanut-almond mixture needed to create a 5 lb nut mix that is 60% peanuts and 40% almonds. This problem involves algebraic equations and is suitable for Grades 8-10.

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