How do you solve the following system: #3x + 2y = 1, 2x - 3y = 10 #?

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Answer 1 · 2018-04-16T03:19:27

#x=23/13, y=-28/13#

#3x+2y=1# equation (1)
#2x-3y=10# equation (2)

From (2): #2x=3y+10 => x=(3y+10)/2#

Substitute #x=(3y+10)/2# into (1);

#3((3y+10)/2)+2y=1#

#3(3y+10)+4y=2#

#9y+30+4y=2#

#13y=-28#

#y=-28/13#

Substitute #y=-28/13# into #x=(3y+10)/2#:

#x=(3(-28/13)+10)/2=23/13#