How do you solve #x+3y=-1#, #y=x+1#?

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Answer 1 · 2018-03-03T07:23:45

For a problem like this you would use substitution method.

Since the problem already gives you the value of #y# (which is #x + 1#), you would use that information and apply it to the equation

#x + 3y = -1#

How you would do this is by replacing #x + 1# (the value of #y#) into the original equation wherever you see #y#.

So:

#x + 3y = -1 #

#x + 3(x + 1) = -1 #

(Solve algebraically)

#x + 3x + 3 = -1#

#4x + 3 = -1#

#4x = -4#

#x = -1 #

Now to find #y# as a number, you would plug in #-1# as #x# and solve:

#y = x + 1#

#y = -1 + 1#

#y = 0 #