Question 859007
Set up a system of three equations with three unknowns.
new brand of plant fertilizer is to be made from three different types of chemicals (A, B, C).
A + B + C = 600

The mixture includes 80% of chemicals A and B.
A + B = 600*0.8
A + B = 480


Chemical B and C must be in ratio of 3 to 4 by weight.
B = (3/4)C



Now plug in the value of any variable from one equation to the other two, in the place of that variable. Say, from the last equation.
We will plug in (3/4)C in both the top two equations.

A + B + C = 600
A + (3/4)C + C = 600
A + (7/4/)C = 600


A + B = 480
A + (3/4)C = 480


Now we have the two equations with two variables (B was eliminated):
A + (7/4)C = 600
A + (3/4)C = 480


From the first equation:
A = 600 - (7/4)C
Plug this into the second equation:
600 - (7/4)C + (3/4)C = 480
Now we collect like terms and:
-(4/4)C = -120
C = 120


Now we have that it takes 120 kg of chemical C for the mixture.
Next, we solve for A and B.


Now you know C, so from the equation:
B = (3/4)C
B = 3/4 * 120
B = 90


And finally, solve for A.
A + B = 480
A = 480 - B
A = 480 - 90
A = 390


It takes 390 kg of chemical A, 90 kg of B and 120 kg of C.


We also want to check.
They must add to 600.
390 + 90 + 120 = 600
A and B together must make 80% of 600.
(390 + 90)/600 = 0.8
And B must equal 3/4 of C
90 * (3/4) = 120.


Great. Hope that helps:)