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Help solving a rational inequality

arl2267

New member
Nov 6, 2012
15
Solve the rational inequality (a-5)/(a+2) < -1.


This is what I got so far:

a-5/a+2 = -1
a-5= -a-2
0= -2a+3

subtract 3 from both sides:

-3=-2a

Divide by -2

3/2=a

I know that the answer is (-2, 3/2), but I'm not sure where the -2 in the answer comes from. Thanks!
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You don't want to multiply through by an expression containing the variable, since it may be either negative or positive. I recommend proceeding as follows:

$\displaystyle \frac{a-5}{a+2}<-1$

Add through by 1:

$\displaystyle \frac{a-5}{a+2}+1<0$

Combine on the left:

$\displaystyle \frac{2a-3}{a+2}<0$

Now, this gives you two critical numbers obtained from the roots of the numerator and denominator. The sign of the expression can only change across these points.

They are: $\displaystyle a=-2,\,\frac{3}{2}$

Now, on a number line mark these points, giving you 3 intervals to test, so pick a test point from within each interval and then check the sign of the expression at these test points. If the sign of the expression is negative, then that interval is part of the solution. Since the inequality is strict, all intervals will be open.
 
Last edited:

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
(a-5)/(a+2) < -1


This is what I got so far:

a-5/a+2 = -1
a-5= -a-2
0= -2a+3

subtract 3 from both sides:

-3=-2a

Divide by -2

3/2=a

I know that the answer is (-2, 3/2), but I'm not sure where the -2 in the answer comes from. Thanks!
You have to be wary with inequalities because it changes direction if you multiply or divide by a negative so ideally we don't want to multiply across the inequality.

$\dfrac{a-5}{a+2} < -1$

$\dfrac{a-5}{a+2} + \dfrac{a+2}{a+2} < 0 $

$\dfrac{2a-3}{a+2} < 0$

For the LHS to be negative the numerator and denominator must have different sign. Points to check are $\frac{3}{2}$ and $-2$ leading to the intervals
$ a < -2 $
$-2 < a < \frac{3}{2}$
$a > \frac{3}{2}$