How do you solve #(1/2)x - (1/3)y = -3# and #(1/8)x + (1/6)y = 0#?

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Answer 1 · 2016-01-18T22:45:06

#(x,y)=(-4,3)#

#(1/2)x-(1/3)y=-3# and #(1/8)x+(1/6)y=0#

The minimum common multiple between 2,3,6 and 8 is 24. So
let's convert all these fractions in something/24:

#(12/24)x-(8/24)y=-(72/24)# and #(3/24)x+(4/24)y=0#

Then multiply all the equalities by 24:

#12 x - 8y =-72# and # 3 x +4 y =0 #

If # 3 x + 4 y =0 # the double #6 x + 8 y# is also 0, so we can
add it to the other equation without change the result:

#6 x + 8 y=0# plus #12 x - 8y =-72# gives #18x=-72#

So, #x=-72/18=-4#

and applying the value #x=-4# to the expression:

# 3 x +4 y =0 #

gives

#3*(-4)+4y=0#
#4y=12#
#y=3#