How do you solve #(x-1)/(2x+3) <=(1)#?

2 Answers
Jun 29, 2015

Answer:

The solution set is: #(-oo, -4] uu (-3/2, oo)#. Here is a method for solving inequalities like this (Without a graphing calculator.)

Explanation:

#(x-1)/(2x+3) <= 1#

Start by comparing some function (expression) to zero.
[Note: this is not really necessary, but is simplifies the question to be asked later when is the expression of the left less that the number on the right. If the number on the right is #0#, then the expression is less when it is negative.]

#(x-1)/(2x+3) -1 <= 0#

Combine the expression to get a single polynomial or rational expression. (In this question, it will be a rational expression.)

#(x-1)/(2x+3) - (2x+3)/(2x+3) <= 0#

#((x-1) - (2x+3))/(2x+3) <= 0#

#(x-1 - 2x - 3)/(2x+3) <= 0#

#(-x-4)/(2x+3) <= 0#

#(-(x+4))/(2x+3) <= 0#

Now, if the expression on the left is going to change from positive (#>0#) to negative (#<0#) or vice versa, then one of two things must happen: either the expression is equal to #0#, of the expression is not defined (because the denominator is #0#).

The number that make the left side #0# or not defined will cut the number line into pieces.
(These numbers are sometimes called "Partition Numbers" because they partition the number line. Sometimes they are called "Key Numbers".)

#(-(x+4))/(2x+3) = 0# #color(white)"xx"# or #color(white)"xx"# #(-(x+4))/(2x+3) # is not defined.

#-(x+4) = 0# #color(white)"xx"# or #color(white)"xx"# #2x+3 = 0#

#x = -4# #color(white)"xx"# or #color(white)"xx"# #x = -3/2#

The only places that the expression on the left might change sign are #-4# and #-3/2#, now we'll see if it does and where it is negative.

The partitions of the number line are (the intervals):

#(-oo, -4)# and #(-4, -3/2)# and #(-3/2, oo)#.

We need to test each interval. There are a few ways to do this, but for many students, the simplest is to pick a test number in each interval and determine the sign of the expression.

#{: ("interval","test number", "result", "pos or neg"), ((-oo, -4), color(white)"xx" -10, (-(-6))/(-17), color(white)"xx" "negative"), ((-4, -3/2), color(white)"xxx" -3, (-(1))/(-1), " positive"), ((-3/2, 00), color(white)"xxx" 10, color(white)"x"(-(14))/23, " negative") :}#

To solve: #(-(x+4))/(2x+3) <= 0#, we want the values of #x# that make the expression negaitve or zero (but not the one that make is undefined).

The solutions are all the numbers in #(-oo, -4] uu (-3/2, oo)#.

Jun 29, 2015

Answer:

Using graphing technology:

Explanation:

To find the solutions of #(x-1)/(2x+3) <= 1# it is easiest to graph the equation #y = (x-1)/(2x+3) -1# and look for the values of #x# where #y <0# (where the graph is below the #x# axis#)

graph{y = (x-1)/(2x+3) -1 [-10.65, 9.36, -4.26, 5.74]}

You can zoom in with a mouse wheel and you can hold and drag the graph around as needed.

It is, I think, fairly clear that the graph is below the #x# axis, so #y < 0# for all #x < -4#

It is also clear that from #x= -4# to some other negative #x#, the graph is above the axis. And past that other #x#, the graph is below again.

That other negative #x# value is #x=-3/2#. For #x> -3/2#, the graph is below the #x# axis.

So the solutions are all values of #x# less than or equal to #-4# (remember that we want to allow #=# -- that is, the inequality is not strict) together with all values of #x# greater than (but not equal to) #-3/2#