How do you graph and solve #|2x+3| <= 7#?

1 Answer
Dec 4, 2015

Answer:

#-5 <= x < 2#

#x in [-5, 2]#

Explanation:

Graphing

Let's start with graphing.

You can graph #|2x+3|# first. How do you do this?

The graph of #|x|# looks like this:

graph{|x| [-15, 15, -3, 12]}

To graph #|2x + 3|#, we need to adapt the slope of the two "arms" and to shift the "elbow" appropriately.

  • the slope is being described by the factor in front of #x#. Here, the slope is #2#.
  • to find the "elbow", set #2x + 3 = 0# and solve for #x#. Thus, the "elbow" is at #(-3/2, 0)#

Thus, the graph of #|2x+3|# is:

graph{|2x+3| [-15, 15, -3, 12]}

Now, the solution of the equation are all #x# values for which the graph is underneath the line #y = 7#:

graph{(y - |2x+3|)(y-7) = 0 [-15, 15, -3, 12]}

So, you already see on the graph that the solution is #-5 <= x < 2# or #x in [-5, 2]#.

How to find the same solution without the graph?

Solving

To solve the inequality, you need to evaluate the absolute value #|2x+3|# .

Generally, for an absolute value, you have

#|x| = { (x, x >= 0), (-x, x < 0):}#

In this case, you need to know for which #x# the equation #2x + 3 = 0# has a solution.
We already know that this is the case for #x = - 3 /2#.

So, we have

#|2x+3| = { (2x+3, x >= -3/2), (-(2x+3), x < -3/2):}#

Thus, we need to consider the two cases:

1) #x >= -3/2#:

#2x + 3 <= 7#

#<=> x <= 2 #

Don't forget to take a look at the condition #x >= -3/2# which needs to hold at the same time.
Here, the solution is #-3/2 <= x <= 2# or #x in [-3/2; 2]#

2) #x < -3/2#:

#-(2x + 3) <= 7#

#<=> -2x - 3 <= 7 #

#<=> -2x <= 10 #

... divide by #-2# and don't forget to flip the inequality sign (this needs to be done if multiplying with or dividing by a negative number)...

#<=> x >= -5#

Don't forget to take a look at the condition #x < -3/2# which needs to hold at the same time.
Thus, the solution for this case is #-5 <= x < -3/2# or #x in [-5; -3/2)#

In total, we have #x in [-3/2; 2]# or #x in [-5; -3/2)# which can be combined to

#x in [-5; 2]#