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

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.

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

Answer 3

To solve this problem, we can set up a system of equations to represent the given information.

Let's denote the number of deluxe radios, standard radios, and economy radios as D, S, and E respectively.

Based on the information provided, we can create the following equations:

Equation 1: 2P(D) + 2P(S) + P(E) = 2200 (total P transistors)
Equation 2: 7Q(D) + 3Q(S) + 2Q(E) = 3400 (total Q transistors)
Equation 3: 1R(D) + 1R(S) + 2R(E) = 1400 (total R transistors)

Now, we can solve this system of equations to find the values of D, S, and E.

First, let's solve Equation 1 for D:
2P(D) + 2P(S) + P(E) = 2200
2P(D) = 2200 - 2P(S) - P(E)
D = (2200 - 2P(S) - P(E)) / 2P

Similarly, we can solve Equation 2 and 3 for S and E respectively:
S = (3400 - 7Q(D) - 2Q(E)) / 2Q
E = (1400 - R(D) - R(S)) / 2R

Now, let's substitute these expressions for D, S, and E back into the original equations to find the values of P, Q, and R.

Substituting D, S, and E into Equation 1:
2P((2200 - 2P(S) - P(E)) / 2P) + 2P(S) + P(E) = 2200
Simplifying, we get:
(2200 - 2P(S) - P(E)) + 2P(S) + P(E) = 2200
2200 = 2200

This equation shows that P can be any value since it cancels out on both sides. Therefore, we cannot determine the value of P.

Similarly, we can substitute D, S, and E into Equation 2 and 3 to find the values of Q and R. However, both equations will result in statements that are always true, meaning Q and R can also be any value.

Given this information, we cannot determine the exact number of radios of each type that can be constructed with the given transistor quantities. However, we know that there are multiple possible combinations of D, S, and E that satisfy the given conditions.