Question 279132
In the first group of equations we can see that the difference between the first and second equations is that 2Z has been added but it makes no difference to the total. So here we can conclude Z=0.


We now have X+Y=5 and X+3Y=11 since Z=0


From these two equations we can see that by adding an extra 2Y to the left (from Y to 3Y) we increase its value by 6 (from 5 to 11). So 2Y=6; giving Y=3.


X+3=5 gives X=2


So X=2; Y=3; Z=0 we verify the following:

2 + 3 + 2x0 = 5

2 + 3 = 5

2 + 3x3 + 0 = 11


The second set of equations is really the same as the first - all equations are multiples of the first set.

Equation1. X+Y+2Z=5 Double it 2X+2Y+4Z=10

Equation2. X+Y=5 Treble it 3X+3Y=15

Equation3. X+3Y+Z=11 Quadruple it 4X+12Y+4Z=44