A manufacturer produces three types of radios: deluxe, standard and economy. Each radio uses three different types of transistors: P, Q and R. The deluxe radio uses 2 P's, 7 Q's and 1 R. The standard contains 2 P's, 3 Q's and 1 R, and the economy model requires 1 P, 2 Q's and 2 R's.

Question

How many radios of each type can be constructed if the total number of transistors (P's, Q's and R's) available are 2200, 3400 and 1400 respectively and all transistors must be used?

Answer 1

number of deluxe : x
number of standard : y
number of econony : z

your system is
2x + 2y + z = 2200
7x + 3y + 2z = 3400
x + y + 2z = 1400

2nd - 3rd : 6x + 2y = 2000 or ---> 3x + y = 1000

double the 1st, subtract the 3rd
3x+3y=3000

subtract the x,y equations:

2y = 2000
y = 1000
sub into 3x+y=1000
3x + 1000= 1000
3x=0
x=0 , !!!!!????

Unless I made some arithmetic error, I think your data is flawed.

Answer 2

I get the same results, x is indeed zero.

See:
http://www.jiskha.com/display.cgi?id=1310358660