An automobile repair shop wants to mix a solution that is 35% pure antifreeze with another solution that is 75% pure antifreeze. How many liters of each solution must be used in other to produce 80 liters of solution that is 50% pure antifreeze?

Answer

To solve the problem of mixing two antifreeze solutions, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = liters of the 35% antifreeze solution
- \( y \) = liters of the 75% antifreeze solution

### Step 2: Set Up Equations
We know that the total volume of the mixed solution should be 80 liters. Therefore, we can write the first equation as:
\[ x + y = 80 \]

Next, we need to account for the concentration of antifreeze in the final solution. The total amount of pure antifreeze in the mixture should equal the sum of the pure antifreeze from each solution. This gives us the second equation:
\[ 0.35x + 0.75y = 0.5 \times 80 \]

### Step 3: Simplify the Second Equation
Calculating the right side of the second equation:
\[ 0.5 \times 80 = 40 \]
So the second equation becomes:
\[ 0.35x + 0.75y = 40 \]

### Step 4: Substitute for \( y \)
From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 80 - x \]

### Step 5: Substitute \( y \) into the Second Equation
Now substitute \( y \) in the second equation:
\[ 0.35x + 0.75(80 - x) = 40 \]

### Step 6: Distribute and Combine Like Terms
Distributing \( 0.75 \):
\[ 0.35x + 60 - 0.75x = 40 \]
Combine like terms:
\[ -0.40x + 60 = 40 \]

### Step 7: Solve for \( x \)
Subtract 60 from both sides:
\[ -0.40x = 40 - 60 \]
\[ -0.40x = -20 \]
Now divide by -0.40:
\[ x = \frac{-20}{-0.40} = 50 \]

### Step 8: Find \( y \)
Now substitute \( x \) back into the equation for \( y \):
\[ y = 80 - x = 80 - 50 = 30 \]

### Conclusion
The automobile repair shop must use:
- **50 liters of the 35% pure antifreeze solution**
- **30 liters of the 75% pure antifreeze solution**