SOLUTION: Ref. Question 1046126: .95(1) = .40(x + 1) .95 = .40x + 40 .95 - 40 = .40x + 40 - 40 .55 = .40x .55 / .40 = .40x / .40 1.375 = x Where am I correct ?

Question 1046146: Ref. Question 1046126:
.95(1) = .40(x + 1)
.95 = .40x + 40
.95 - 40 = .40x + 40 - 40
.55 = .40x
.55 / .40 = .40x / .40
1.375 = x
Where am I correct ?
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
A 1 qt. mixture of alcohol and water is 95% alcohol. The mixture is to be reduced to 40% alcohol. Determine amount of distilled water to be added.

A clear way to think of it is ;
and your variable will be amountAddWater or amount of water to add to the mixture.

, which is written using decimal fractions instead of percents. The amount of water to add is assigned as v.

SOLVE this equation for v.

-
There is a decimal mistake in the steps you asked about. Stepping through mine just two ,

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