SOLUTION: Good evening. I am taking a college level business calculus class. We are currently reviewing algebra. I am stuck on one problem. It is: A-5 ---- < -1 A+2 I know t

Algebra ->  Inequalities -> SOLUTION: Good evening. I am taking a college level business calculus class. We are currently reviewing algebra. I am stuck on one problem. It is: A-5 ---- < -1 A+2 I know t      Log On


   



Question 341756: Good evening. I am taking a college level business calculus class. We are currently reviewing algebra. I am stuck on one problem. It is:
A-5
---- < -1
A+2

I know to take the -1 times A + 2 to get A - 5 < A + 2 and I know to take the +2 and subtract it from A - 5 to get A - 5 - 2, but what happens to the A on the right side. Is it not supposed to be a 0? I am really confused and lost and, therefore, unable to continue because the rest of this section contains similar problems. Please help me figure this problem out.

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A-5
---- < -1
A+2
---------------------
The boundary of the inequality is (A-5)/(A+2) = -1
-------------------
Since the denominator cannot be zero, A cannot be -2.
---
The numerator cannot be zero so A cannot be 5
-------------------------------
To solve the inequality:
1: Draw a number line
2: Plot the values A = -2 and A = 5
3: Check a test value from each of the resulting 3 intervals
to see where the solutions are for the inequality.
----
A-5
---- < -1
A+2
Check A = -10; (-10-5)/(-10+2) < -1, false so no solutions in (-inf,-2)
Check A = 0; (0-5)/(0+2)< -1, true so solutions in (-2,5)
Check A = 10; (10-5)/(10+2) < -1, false so no solutions in (5,+inf)
====
Final solution: -2< A < 5
============
Graph:
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%28x-5%29%2F%28x%2B2%29%2C-1%29
============
Cheers,
Stan H.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

The other tutor's solution is incorrect. It should be

-2%3Cx%3C3%2F2

Good evening. I am taking a college level business calculus class. We are currently reviewing algebra. I am stuck on one problem. It is:
A-5
---- < -1
A+2

I know to take the -1 times A + 2 to get A - 5 < A + 2

There's one of your errors! -1 times A + 2 is -1(A + 2) and then you
use the distributive principle to remove the parentheses, and
you end up with -A - 2, (not A + 2).

But also when you multiply through by a variable quantity, A + 2, if that
quantity happens to be negative you would have to change the direction
of the inequality.  So 

%28A-5%29%2F%28A%2B2%29%3C-1

would only become

A-5%3C-A-2 if A%2B2%3E0 which is the same as A%3E-2

But it would become

A-5%3E-A-2 if A%2B2%3C0 which is the same as A%3C-2

So two cases must be considered to which if either is valid of
if they both are:

Solving the first case

Case 1:

A-5%3C-A-2 if A%2B2%3E0 or A%3E-2

Add A to both sides:

A-5%2BA%3C-A-2%2BA if A%3E-2

2A-5%3C-2 if A%3E-2

Add 5 to both sides

2A-5%2B5%3C-2%2B5 if A%3E-2

2A%3C3 if A%3E-2 

Divide both side by 2

A%3C3%2F2 if A%3E-2

That gives us the solution set:

%22%28-2%2C%223%2F2%22%29%22

--------------------

Solving Case 2:

A-5%3E-A-2 if A%2B2%3C0 or A%3C-2

Add A to both sides:

A-5%2BA%3E-A-2%2BA if A%3C-2

2A-5%3E-2 if A%3C-2

Add 5 to both sides

2A-5%2B5%3E-2%2B5 if A%3C-2

2A%3E3 if A%3C-2 

Divide both side by 2

A%3E3%2F2 if A%3C-2

This is a contradiction, since A cannot be greater than 3%2F2
and at the same time be less than or equal to -2, so the only solution
is given by case 1.

%22%28-2%2C%223%2F2%22%29%22

Edwin