How do you solve #4|7-y|-1=11#?

1 Answer
Jan 16, 2017

Answer:

See the entire solution process below:

Explanation:

First, we need to go through the necessary steps to isolate the absolute value term:

#4abs(7 - y) - 1 + color(red)(1) = 11 + color(red)(1)#

#4abs(7 - y) - 0 = 12#

#4abs(7 - y) = 12#

#(4abs(7 - y))/color(red)(4) = 12/color(red)(4)#

#(color(red)(cancel(color(black)(4)))abs(7 - y))/cancel(color(red)(4)) = 3#

#abs(7 - y) = 3#

Now we can solve for #y#. Because the absolute value converts any negative or positive term into it's positive form we need to solve the term within the absolute value for both the positive and negative form of what it is equated to.

Solution 1)

#7 - y = 3#

#7 - y + color(red)(y) - color(blue)(3) = 3 + color(red)(y) - color(blue)(3)#

#7 - color(blue)(3) - y + color(red)(y) = 3 - color(blue)(3) + color(red)(y)#

#7 - color(blue)(3) - 0 = 0 + color(red)(y)#

#4 = y#

#y = 4#

Solution 2)

#7 - y = -3#

#7 - y + color(red)(y) + color(blue)(3) = -3 + color(red)(y) + color(blue)(3)#

#7 + color(blue)(3) - y + color(red)(y) = -3 + color(blue)(3) + color(red)(y)#

#7 + color(blue)(3) - 0 = 0 + color(red)(y)#

#10 = y#

#y = 10#

The solution is #y = 4# and #y = 10#