How do you solve the system of equations #x+ 3y = - 11# and #4x + 5y = - 16#?

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Answer 1 · 2018-01-27T03:23:27

#x=1# and #y = -4#

Let's use substitution:

#x=-11-3y#

#4(-11-3y)+5y = -16#

#-44-12y+5y = -16#

#28 = -7y#

#y = -4#

Now let's solve for #x#:

#x = -11-3(-4)#

#x = -11+12#

#x=1#

Now, to double check our work, let;s substitute #-4# and #1# for #y# and #x# and see if we still get the same answer:

#4(1)+5(-4)# should equal #-16#

#4-20#

#-16=-16#

We were right!