How do you solve #y=|x|# and #y=x+2# using substitution?

1 Answer
Jul 11, 2017

Answer:

#(-1,1)#

Explanation:

Use the substitution #y = absx#

#y = x+2#

#absx = x + 2#

Now subtract #x# from both sides.

#absx - x = 2#

Hmm... how do we simplify this? Well, we have three cases:

  1. #" "x# is positive
  2. #" "x# is zero
  3. #" "x# is negative

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If #x# is positive, then #absx = x#, so we can use this substitution:

#absx - x = 2, " " x >0#
#x-x =2, " "color(white). x>0#
#0 = 2, " "" "" "x>0#

And since #0# is never equal to #2#, this means we have no positive solutions.

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Next, let's try case 2, when #x=0#.

#|0| - 0 = 2#

#0 = 2#

Again, #0# cannot equal #2#, so our solution cannot be #x=0#.

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Finally, when #x# is negative, #absx# is the same as #-x#, since to make a negative number positive, you add another negative sign. (Two negatives make a positive!) So, we can use the substitution #absx = -x#:

#absx - x = 2, " " x<0#
#-x-x = 2, " " x<0#
#-2x = 2, " "" " x<0#

Now we can divide both sides by #-2# to find #x#:

#(-2x)/(-2) = 2/(-2), " "x<0#

#x = -1#

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So this is the #x# coordinate for our solution. To find the #y# coordinate, we simply plug #x=-1# back into one of the original equations and solve for #y#.

#y = absx#
#y = abs(-1#
#y = 1#

So #x=-1# and #y=1#. Therefore, our solution is #(-1,1)#.

Final Answer