Question

HELP ME PLEASE.
one acetic acid solution is 70% water, and another 30% water. how many liters of each solution should be mixed to produce 20 liters of a solution that is 40% water? solve this by using a system of equation.

Answer 1

I misread the question...

Answer 2

\(\large \color{black}{\begin{align}
\normalsize \text{u have two equations}\\~\\
\dfrac{70-40}{40-30}&=\dfrac{y}{x}........(1) \\~\\
x+y&=20............(2)
\end{align}}\)

Answer 3

http://www.purplemath.com/modules/mixture.htm

Answer 4

20 liters of 40% water is 8 gal of water. You will need 8 gal of water to obtain this mixture. This 8 gal of water is going to come from the 2 mixtures of which one is 70% water and the other is 30% water.

Answer 5

Let x equal the amount of 70% water solution.
20 - x will then equal the amount of the other solution (30% water) then the actual water from each of these solutions would be (as given by the problem)
.7x + .3(20 - x) = 8
.7x + 6 - .3x = 8

Answer 6

Solve for x
Solve for 20 - x

Answer 7

@katchen_97 Did you follow with understanding?