Solving a percent mixture problem using a system of linear equations

Question

I am struggling with these type of equations and I need help

A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 

60%

 salt and Solution B is 

80%

salt. She wants to obtain 

120

 ounces of a mixture that is 

75%

 salt. How many ounces of each solution should she use?

Answer 1

Let A = ounces of Solution A and B = ounces of Solution B.

How much salt is in Solution A? 0.6A

How much salt is in Solution B? 0.8B

She wants 120 ounces, so A + B = 120. (Remember, A and B are ounces.)

That 120 ounces should be 75% salt, so

0.75 salt/ounce * 120 ounces = 90 salt (That's total salt in the 120 ounces)

0.6A + 0.8B = 90 (That's the amount of salt from ounces of A plus the amount of salt from ounces of B equals that amount of salt she wants in the mixture.)

Now, you have 2 equations with 2 variables.

A + B = 120

0.6A + 0.8B = 90

Rearrange the first equation.

B = 120 - A

Substitute into the second equation.

0.6A + 0.8(120 - A) = 90

Distribute the 0.8.

0.6A + 96 - 0.8A = 90

Combine like terms.

96 - 0.2A = 90

Subtract 96 from each side.

-0.2A = -6

Divide each side by -0.2.

A = 30

Substitute that answer into the first equation.

30 + B = 120

Subtract 30 from each side.

B = 90

Remember A and B stood for ounces of each solution, so she needs 30 ounces of solution A and 90 ounces of solution B.

Note: You also could've thought 75 is 3/4 of the way from 60 to 80, so 3/4 of the mixture is solution B and 1/4 of the mixture is solution A.

The mixture is 120 ounces.

3/4 of 120 is 90.

1/4 of 120 is 30.

She needs 90 ounces of solution B and 30 ounces of solution A.