How do you solve the following system?: #-x -6y =-1, 3x -y = -4#

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Answer 1 · 2015-11-17T21:52:01

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Isolate one variable in one the equations. Then use the resulting equivalent to replace the same variable in the other equation

#-x - 6y = -1#
#3x - y = -4#

If we isolate #x# in the first equation, we will have

#-x - 6y = -1#
#=> x = -6y + 1#

Now, let's replace #x# in the second equation

#3x - y = -4#
#=> 3(-6y + 1) - y = -4#
#=> -18y + 3 - y = -4#
#=> -19y + 3 = -4#
#=> -19y = -7#
#=> y = 7/19#

Now that we have the value for #y#, substitute it into any of the equalities above to get #x#

#x = -6y + 1#
#=> x = -6(7/19) + 1#
#=> x = -42/19 + 1#

#=> x = (-42 + 19)/19#

#=> x = -23/19#