How do you solve #abs(2x+1)=5#?

2 Answers
Nov 7, 2017

Answer:

See a solution process below:

Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1:

#2x + 1 = -5#

#2x + 1 - color(red)(1) = -5 - color(red)(1)#

#2x + 0 = -6#

#2x = -6#

#(2x)/color(red)(2) = -6/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -3#

#x = -3#

Solution 2:

#2x + 1 = 5#

#2x + 1 - color(red)(1) = 5 - color(red)(1)#

#2x + 0 = 4#

#2x = 4#

#(2x)/color(red)(2) = 4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 2#

#x = 2#

The Solutions Are: #x = -3# and #x = 2#

Nov 7, 2017

Answer:

#x = {2,-3}#

Explanation:

We can tackle this by considering how #|a| = |-a|#

So hence;

# |-(2x+1)| = |2x+1| = 5#

So hence, #2x+1 = 5#
But also #-(2x+1) = 5#

As # |-(2x+1)| = |2x+1|#

So hence solving both linear equations we yield;

#x = {2,-3}#