Question 1059451: pure alcohol is added to 50 gallons of coolant mixture that starts out 40% alcohol. Write and expression for the percentage of alcohol in the mixture (y) in terms of gallons of alcohol added (x). I received help with the expression which was x+.4(50)=y(x+50), but i need the explanation for this equation.
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! So, we don't know how much alcohol was added to the mixture. Let's call the amount of alcohol x. And 1x, or simply x is pure alcohol. Now, 0.40(50) is the quantity at 40% and x+0.40(50) is when we add the pure alcohol (x) to the mixture.
:
Next, we know that a quantity of alcohol x was added to 50 gallons. So we know the final quantity will be x+50. But what will be the concentration in that final mixture? We don't know. Let's call the concentration in the final
mixture y. Put it all together:
:
x+0.40(50) = y(x+50)
email if you have questions or need more help.
John
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