How do you solve the system #0.4x-0.1y=2, 0.2x+0.5y=1#?
1 Answer
Jan 14, 2017
See explanation.
Explanation:
The starting system is:
#{ (0.4x-0.1y=2),(0.2x+0.5y=1):}#
If we multiply the second equation by
#{ (0.4x-0.1y=2),(-0.4x-y=-2):}#
Now we can add both sides of the equations to get an euation of 1 variable:
#-1.1y=0#
#y=0#
Now we have to substitute the calculated value of
#0.4x=2#
#x=2/0.4=20/4=5#
Answer:
The solution of this systemm is: