Question

Solve the following equations by the method of
reduction:
x + 3y + 3z= 12
x + 4y + 4z = 15
x + 3y + 4z= 13​

Answer 1

Answer:

» x + 3y + 3z = 12 ... 1

» x + 4y + 4z = 15 ... 2

» x + 3y + 4z = 13 ... 3

from 3

» x + 3y + 3z + z = 13

and from 1

» 12 + z = 13

Therefore, z = 1 ... 4

from 2

» x + 3y + 3z + y + z = 15

from 1 and 4

» y = 1 ... 5

from 1, 4 and 5

» x + 3y + 3z = 12

» x = 12 - 6 = 6