SOLUTION: How many liters of a 20 percent solution of acid should be mixed with 50 percent solution of acid to obtain a 9 liters of 30 percent solution? Need a equation to be able to solv

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Question 1001538: How many liters of a 20 percent solution of acid should be mixed with 50 percent solution of acid to obtain a 9 liters of 30 percent solution?
Need a equation to be able to solve
I need to solve this in 3 ways:
1) substitution
2) linear combination
3) graphing
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
You can use ONE equation with ONE variable this way:
Let v= how much of 20% to use; and then 9-v is how much of the 50% to use.
%2820v%2B%289-v%2950%29%2F9=30
Solve for v, and substitute for v to evaluate 9-v.
Maybe this all will satisfy your (1) substitution and (2)....No, just (1).

(2) Linear Combination might be taken to mean part of Linear Algebra, but you might be able to at least make the system:
x, and y, for how much 20% and 30% respectively.
system%28x%2By=9%2C%2820x%2B50y%29%2F9=30%29, and then arrange what you need from this system.

(3) Graphing.
I will not show it all here, but your (2) material will give the two equations to graph. Look for the intersection. Be sure to simplify both equations and solve each for y in terms of x, and then use the slope-intercept equations for help in graphing the two lines.