How do you solve the system #3x+2y =3#, #2x+0y = -5#?

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Answer 1 · 2015-11-15T14:11:04

#x = -5/2#
#y = 21/4#

We can already get the value of #x# in the second equation.

#2x + 0y = -5#

#=> 2x = -5#

#=> x = -5/2#

Now, to get the value of #y#, we plug-in the value of #x# in the first equation

#3x + 2y = 3#

#=> 3(-5/2) + 2y = 3#

#=> -15/2 + 2y = 3#

#=> 2y = 3 + 15/2#

#=> 2y = (6 + 15)/2#

#=> 2y = 21/2#

#=> y = 21/4#