How do you solve this system of linear equations: 3x-2y=8 and 2x-3y=-6?

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Answer 1 · 2015-12-06T09:57:49

#x =36/5#, #y=34/5#

Multiply the first equation by #2# and the second one by #3#:

#6x-4y=16#
#6x-9y=-18#

Subtract the equations:

#5y = 34 \implies y=34/5#

Now that #y# is known, we can deduce #x# from any of the two equations:

#6x-4y=16 \implies 6x-4*34/5 = 16 \implies 6x-136/5 = 16#

So, #x = (16+136/5)/6 = 36/5#