How do you solve the system #1/3 x - 1/12 y = -7# and #1/3 x + 1/6 y = 11#?

2 Answers

Answer:

x = -3 and y = 72

Explanation:

So we are given with #2# equations

equation 1 is #1/3x - 1/12y# = -7

equation 2 is #1/3x + 1/6y# = 11

equation 1 can be written as #4x - y = -84 #
(multiplied the whole equation by 12)

equation 2 can be written as #2x + y = 66#
(multiplied whole equation by 6)

now we add both the equations

==> #4x - y + 2x + y = -84 + 66#
==> #6x = -18#

==> #x = -18/6#

==> #x = -3#

now we substitute #x = -3# in any one of the equations
we will get #y = 72#

Mar 19, 2018

Answer:

#x = -3 and y =72#

Explanation:

Both equations have the same #x#- coeffiecients.

Rewrite both equations:

#1/3x = 1/12y -7" "and" "1/3x= -1/6y +11#

#1/3x = 1/3x# , so equate the right sides of the equations.

#1/12y -7 = -1/6y +11" "larr# multiply by #12#

#y-84 = -2y+132#

#y +2y = 132+84#

#3y = 216#

#y = 72#

Substitute to find #x#

#1/3x = 1/12 (72)-7#

#1/3x = 6-7#

#1/3x = -1#

#x =-3#