3) how many liters of a 20% acid solution must be mixed with a 60% solution to obtain 40 liters of a 35% solution? (25,15)thats the answer i need the steps

let x = the number of liters of the 20% solution. 

let y = the number of liters of the 60% solution. 

you want x + y to be equal to 40 liters. 

x is the number of liters total in the first solution.
y is the number of liters total in the second solution. 

you want .2 * x + .6 * y to be equal to .35 * 40 

.2 * x is the number of liters of acid in the first solution.
.6 * y is the number of liters of acid in the second solution.
.35 * 40 is the number of liters of acid in the final solution. 

you have two equations that need to be solved simultaneously. 

they are: 

x + y = 40
.2x + .6y = .35*40 

simplify these equations to get: 

x + y = 40
.2x + .6y = 14 

you can solve by substitution or by elimination or by graphing.
i will solve this one by graphing.
this means to graph both equations and find the intersection. 

the graph looks like this: 

 
the graph says the intersection is at the coordinate point of (25,15). 

this means that x = 25 and y = 15. 

x is the number of liters of the 20% solution.
y is the number of liters of the 60% solution. 

the formula of .2x + .6y = 14 becomes .2 * 25 + .6 * 15 = 14.
simplify this equation to get 14 = 14.
this confirms the solution is good. 

The answer is 14

Have a great night and best of luck my friend :)