Challenging Algebra word problem. System of equations.?

Question

This algebraic systems of equations word problem has got me stuck because it includes a ratio in the problem. The book doesn't teach ratios for another 3 chapters! Please illustrate what I must do in order to solve this System of equations word problem.


A plant food is to be made from three chemicals. The mix must include 60% of the first and second chemicals. The second and third chemicals must be in a ratio of 4 to 3 by weight. How much of each chemical is needed to make 750 kg of the plant food?

Answer 1

let the 3 chemicals be a,b and c respectively.
eq1: 60% of 750 is a and b which is
0.6(750) = a + b ------------in symbols
450 = a + b

eq2: the remaining 40% of the 750kg plant food is made up of chemical c.
40% of 750 = c
0.4(750) = c
c = 300

eq3: if the ratio of chemical b to chemical c is 4:3 then the proportion should be
4:3 = b:300 or 4/3 = b/300 in fraction form
cross multiply
3b = 1200 (divide by 3)
b = 400

go back to eq1
a + b = 450
a + 400 = 450
a = 450 - 400
a = 50 // answer

Answer 2

Part 1

60% + 60% + x(unknown of 3rd)= 100% of the food.

y= 1st chemical
m= chemicals 2 and 3

y + 4m + 3m= 750 kg

What is 60% of 750 kg?

Then you have y and 4m.
Subtract that from 750 kg to get 3m.
Place the numbers in a ratio to see if second and third chemicals are in the true proportion.
I hope this helps