How do you solve the system #0.4x-0.1y=2, 0.2x+0.5y=1#?

1 Answer
Jan 14, 2017

Answer:

See explanation.

Explanation:

The starting system is:

#{ (0.4x-0.1y=2),(0.2x+0.5y=1):}#

If we multiply the second equation by #-2# the coefficients of #x# will be opposite numbers:

#{ (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 #y# to calculate #x#:

#0.4x=2#

#x=2/0.4=20/4=5#

Answer:

The solution of this systemm is:

#{(x=5),(y=0):}#