solving a percent mixture using a system of linear equations

Question

A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 70% salt and Solution B is 95% salt. She wants to obtain 50 ounces of a mixture that is 85% salt. How many ounces of each solution should she use?

Answer 1

Solution A = 70% salt

Solution B= 95% salt

Solution C=85% salt with 50 ounces

So,

Solution A + Solution B = Solution C

Change the percentages into decimals in order to solve

0.70x+0.95y=0.85(50) <----Eq(1)

x+y=50 <-----Eq(2)

Solve Eq(2) in terms of x:

x+y=50

y=50-x <---- this we can plug into Eq(1)

Now solve for x after you plug in y=140-x:

0.70x+0.95(50-x)=0.85(50)

0.70x+47.5-x=42.5

-0.30x=-5

x=16.7 ounces <----this can now be plugged into Eq(2) to find the number of ounces in Solution B

Plug in x=16.7 into Eq(2):

x+y=50

16.7+y=50

y=33.3 ounces

Solution A requires 16.7 ounces of 70% salt

Solution B requires 33.3 ounces of 95% salt