How do you solve the system #x/3+y/5=2# and #x/3 - y/2= -1/3#?

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Answer 1 · 2016-03-25T22:22:08

I did this in a hurry so you need to check the calculations!
#y=40/21#

#x=51/4#

My first reaction was to consider getting rid of the fractions. That was, until I spotted the coefficients of #x# are the same in both equation.

Given:

#x/3+y/5=2#..............................(1)

#x/3-y/2=-1/3#...........................(2)

Subtract equation (2) from (1)

#0+y/5+y/2=2+1/3#

#(2y+5y)/10=4/3 #

#7y=40/3#

#y=40/21#

Now all you have to do is substitute

Substitute in equation (1) giving

#x/3+(1/5xx40/21)=2#.........................(1_a)

#x/3 = 2-8/21 = 34/8#

#x=(3xx34)/8 =12 3/4 =51/4#