# Linear Equations

First equation:

A company makes three products X, Y and Z. Each product requires processing by three machines A, B and C. The time required to produce one unit of each product is shown below.

PRODUCT MACHINE A MACHINE B MACHINE C

X 1 2 2

Y 2 8 3

Z 2 1 4

The machines are available for 200, 525 and 350 hours each month. How many units of each product can be manufactured per month if all three machines are utilized to their full capacity?

Second Equation:

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?

Third equation:

The table below shows the number of hours required in each of two departments to make one unit of various products A, B and C. For example, product B requires 1 hour of time in department I and 3 hours in department II.

HOURS REQUIRED PER UNIT OF PRODUCT

DEPARTMENT A B C

I 1 1 9

II 1 3 7

Find the number of units of A, B, and C which could be made if department I has 75 hours available and Department II has 65 hours available. It is necessary that all of the available hours be used.

#### Solution Preview

(1) Using the given information we can create three simultaneous equations, each representing information for different machine, using the relation given below.

X * Machine Hours + Y * Machine Hours + Z * Machine Hours = Total Machine Hours

X + 2Y + 2Z = 200 -- <i> (For Machine A)

2X + 8Y + Z = 525 -- <ii> (For Machine B)

2X + 3Y + 4Z = 350 -- <iii> (For Machine C)

Where X, Y and Z represent the number of units of each product that can be manufactured per month when all three machines are utilized to their full capacity.

Let us solve these equations now.

<ii> - 2*<i> gives

(2X + 8Y + Z) - (2X + 4Y + 4Z) = 525 - 2*200

=> 4Y - 3Z = 125 -- <iv>

<ii> - <iii> gives

(2X + 8Y + Z) - (2X + 3Y + 4Z) = 525 - 350

=> 5Y - 3Z = 175 -- <v>

<v> - <iv> gives

(5Y - 3Z) - (4Y - 3Z) = 175 - 125

=> Y = 50

Putting the value of Y in <iv>, we get

4*50 - 3Z = 125

=> 3Z = 200 - 125

=> Z = 75/3 = 25

Putting the value of Y and Z in <i>, we get

X + 2*50 + 2*25 = 200

=> X + 150 = 200

=> X = 200 - 150 = ...

#### Solution Summary

The solution first explains how simultaneous equations can be formed out of the given relations in each of the questions, and then gives step by step solution to these equations in each case. Note that in third case there will be more than one set of values of (A,B,C) that will satisfy the given relations.