How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifreeze? Use the six-step method. ...?

To solve this mixture problem, you can use the following six-step method:

Step 1: Assign Variables

• Let x
• 
  
• x be the number of gallons of the 80% antifreeze solution needed.

Step 2: Write the Initial Equation

• The total amount of antifreeze in the 80% solution plus the total amount in the 20% solution must equal the total amount in the final mixture.
• 0.80x+0.20(60)=0.70(x+60)
• 
  
• 0.80x+0.20(60)=0.70(x+60)

Step 3: Solve for the Unknown

0.80x+12=0.70x+42

0.80x+12=0.70x+42

Step 4: Simplify the Equation

0.10x=30

0.10x=30

Step 5: Solve for x

x

x=300.10=300

x=0.10

30

​=300

Step 6: Check the Answer

• Ensure that the solution makes sense in the context of the problem. In this case, it means confirming that 300 gallons of the 80% solution mixed with 60 gallons of the 20% solution will indeed result in a mixture that is 70% antifreeze.

Therefore, you would need 300 gallons of the 80% antifreeze solution to mix with 60 gallons of the 20% antifreeze solution to get a mixture that is 70% antifreeze.