Question 697274
If a chemist (like me) wanted to mix a 22% acid solution with a a 50% acid solution to make a 22% acid solution, she/he would first want to know what kind of acid we are dealing with, and how accurate we need to be.
 
However, in math problems (unlike in real life), when you mix volumes of different solutions the final volume is the sum of the original volumes (and nothing dangerous can happen, either).
 
IF YOU HAVE STUDIED systems of linear equations:
Volume balance:
{{{x}}} = volume of 22% acid solution to be used (in L)
{{{y}}} = volume of 50% acid solution to be used (in L)
final volume = {{{x+y}}}L = 28 L (the abbreviation, as per the SI, should be a capital L)
So {{{highlight(x+y=28)}}} <--> {{{highlight(y=28-x)}}} is our first equation.
 
Acid amount balance:
Whatever the units are (grams, liters, or whatever)
{{{x}}} L of 22% solution contain {{{0.22x}}} units of acid
{{{y}}} L of 50% solution contain {{{0.50y}}} units of acid
{{{28}}} L of 30% solution should contain {{{0.30*28=8.4}}} units of acid
So {{{0.22x+0.50y=8.4}}}
Multiplying both sides of the equal sign times {{{50}}}, we get
{{{highlight(11x+25y=420)}}}
 
If you have studied systems of linear equations, you would say that you have the system
{{{system(x+y=28,11x+25y=420)}}}
Then, you could substitute the expression for y, {{{y=28-x}}} into {{{11x+25y=42)}}} and get {{{11x+25(28-x)=420}}} to solve for {{{x}}}.
 
{{{11x+25(28-x)=420}}} --> {{{11x+700-25x=420}}} --> {{{11x+700-25x-70=420-700}}} --> {{{-14x=-280}}} --> {{{(-14x)/(-14)=(-280)/(-14)}}} --> {{{highlight(x=20)}}}
 
IF YOU HAVE NOT STUDIED systems of linear equations:
Volume balance:
{{{x}}} = volume of 22% acid solution to be used (in L)
{{{28-x}}} = volume of 20% acid solution to be used (in L)
The amount of acid in the final mixture would be
{{{0.22x+0.50(28-x)}}}
and that must equal {{{28*0.30=0.84}}} ,
so our equation is
{{{highlight(11x+25(28-x)=420)}}}
and that equation can be solved as shown avbove.