### Math Notes

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#### Algebra Solutions

##### Topics || Problems

How much of a 40% solution of alcohol and how much of an 80% solution should be mixed to give 40 gallons of 50% solution? Solution:
 A(Alcohol), W(water), (M)Mixture A 40% 80% 50% W 60% 20% 50% x y 40 Gallons

$$x+y = 40$$

$$x = 40-y$$

$$0.40x +0.80y = 0.50(40)$$

$$0.40(40-y) +0.80y = 0.50(40)$$

$$16 +0.4 y = 20$$

$$y = 10$$

$$x = 40-10$$

$$x =30$$

Thus, to create a solution a 30 gallons is needed from the 40% solution and 10 gallons from the other.