How do you solve # abs( x+5) = 2x+3# and find any extraneous solutions?

Question

Nathan Axel · Accepted Answer

2 and an extraneous solution, #-8/3#.

#|x+5|=2x+3# is the combined equation for the pair #x+5=2x+3#, when #x>=-5# and #-(x+5)=2x+3#, when #x<=-5# and So, we get solutions, #x=2>-5# and #x=-8/3# not #<-5#, respectively. The second is an extraneous (as it is notnot a) solution.

Kennedy Adams · Answer

undefined To solve $ |x + 5| = 2x + 3 $ and check for extraneous solutions, follow these steps:

1. Split into two cases:
   - Case 1: $ x + 5 \geq 0 $.
   - Case 2: $ x + 5 < 0 $.
2. Solve each case separately:
   - Case 1: $ x + 5 = 2x + 3 $.
     - Solve for $ x $: $ x = 2 $.
   - Case 2: $ -(x + 5) = 2x + 3 $.
     - Solve for $ x $: $ -x - 5 = 2x + 3 $.
     - $ -3x = 8 $.
     - $ x = -\frac{8}{3} $.
3. Check for extraneous solutions:
   - Substitute $ x = 2 $ into the original equation: $ |2 + 5| = 2(2) + 3 $.
     - $ |7| = 7 $ which is true.
   - Substitute $ x = -\frac{8}{3} $ into the original equation: $ |-\frac{8}{3} + 5| = 2\left(-\frac{8}{3}\right) + 3 $.
     - $ \left|-\frac{7}{3}\right| = -\frac{1}{3} $ which is false.
4. Since $ x = -\frac{8}{3} $ is extraneous, the only solution is $ x = 2 $.