Question 472581
A company makes 3 types of cables. Cable A requires 3 black, 3 white, and 2 red wires.
 B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red.
 They used 100 black, 110 white, and 90 red wires.
 How many of each cable were made?:
Let a = no. of A cables
Let b = no. of B cables
Let c = no. of C cables
:
Write an equation for each color wire
black: 3a + b + 2c = 100
white: 3a + 3b + c = 110
red::: 2a + b + 2c = 90
:
Note that using the first and 3rd equation we can eliminate b and c
3a + b + 2c = 100
2a + b + 2c = 90
-------------------subtraction eliminates b and c, find a
a = 10 type A cables
:
replace a with 10 in the 1st equation
3(10) + b + 2c = 100
b + 2c = 100 - 30
b + 2c = 70
b = (70-2c); we can use this form for substitution
:
Do the same with the 2nd equation
3(10) + 3b + c = 110
3b + c = 110 - 30
3b + c = 80
Replace b with (70-2c)
3(70-2c) + c = 80
210 - 6c + c = 80
-6c + c = 80 - 210
-5c = -130
c = {{{(-130)/(-5)}}}
c = +26 C type cables
:
Find b 
b = 70 - 2c
b = 70 - 2(26)
b = 70 - 52
b = 18 B type cables
:
:
Check this in the red equation
 2a + b + 2c = 90
2(10) + 18 + 2(26) = 
20 + 18 + 52 = 90, confirms our solutions of a=10, b=18, c=26