How do you solve #4x - 3y = 22 # and #4x + 5y = 6#?

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Answer 1 · 2017-02-16T12:22:55

#y = -2#

#x = 4#

Set it out in a column like so:

#4x - 3y = 22#
#4x + 5y = 6#

Then subtract the second equation from the first one to get:

#-8y = 16#

which can be simplified to

#-y = 2" "# or #" "y = -2#

This, in turn, can be used to work out the #x# value:

#4x -10 = 6#

#4x = 16#

#x = 4#

Hopefully this helps...