How do you solve #4 + abs(r + 2 ) =7#?

Question

Abigail Allen · Accepted Answer

First subtract #4# from both sides to get: #abs(r+2) = 3#

So either #r+2 = 3#, giving #r = 1#

or #r+2 = -3#, giving #r = -5# Let's look at the definition of absolution value first: #abs(x) = x# if #x > 0# #abs(x) = 0# if #x = 0# #abs(x) = -x# if #x < 0# So if #abs(x) = a#, then #a >= 0# and #x = +-a# Now subtract #4# from both sides of our original equation to get: #abs(r+2) = 3# From what we have found, #r+2 = +-3# Subtract #2# from both sides to get: #r = -2 +- 3# That is #r = -5# or #r = 1#

Eli Assouline · Answer

undefined To solve the equation 4 + |r + 2| = 7, follow these steps:

1. Subtract 4 from both sides to isolate the absolute value term: |r + 2| = 3.
2. Rewrite the equation as two separate equations, one positive and one negative: 
   a. r + 2 = 3
   b. -(r + 2) = 3
3. Solve each equation separately:
   a. For r + 2 = 3, subtract 2 from both sides to find r: r = 1.
   b. For -(r + 2) = 3, distribute the negative sign and then solve for r: -r - 2 = 3. Add 2 to both sides and then multiply by -1 to isolate r: r = -5.
4. Check both solutions by substituting them back into the original equation.