How do you solve the system of equations #3x + 9y = 1# and #- 12x + 3y = 16#?

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Answer 1 · 2017-06-28T04:06:49

#x=-47/39# and #y=20/39#

Lets call #3x+9y=1# the first equation and #-12x+3y=16# the second equation.

First multiply the first equation by #4# so that it has #12# as a coefficient of the #x# term

#4xx(3x+9y)=4xx(1)#

#12x+36y=4#

Comparing to the second equation, we get

#" "12x+36y=4#
#-12x+3y=16#

Adding the two equations to each other gives

#0x+39y=20#

#39y=20#

#y=20/39#

Finally, take #y=20/39# and plug it into either of the first two original equations and solve for #x#

#3x+9color(red)(y)=1#

#3x+9(color(red)(20/39))=1#

#3x+180/39=1#

#3x=1-180/39#

#3x=39/39-180/39#

#3x=-141/39#

Dividing both sides by #3# gives

#x=-47/39#

ANSWER: #x=-47/39# and #y=20/39#