How do you solve the system #0.3x-0.2y=0.5# and #x-2y=-5# using substitution?

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Answer 1 · 2017-06-12T07:50:21

#x=5#, #y=5#

Here we have a system of equations:

#0.3x-0.2y=0.5#
#x-2y=-5#

Let's focus on the second equation and rewrite the equation so that #x# is in terms of #y#:

#x-2y=-5#

#x=2y-5#

Now, we can substitute this equation for #x# into the first equation and solve for #y#:

#0.3(2y-5)-0.2y=0.5#

#0.3(2y)-0.3(5)-0.2y=0.5#

#0.6y-1.5-0.2y=0.5#

#0.6y-0.2y=0.5+1.5#

#0.4y=2#

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

Now that we have found #y#, we can (again) substitute it back into the second equation to find #x#:

#x-2(5)=-5#

#x-10=-5#

#x=5#

So, our answers are:

#x=5# and #y=5#