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Help in inequalities algebra, please! I am loosing my mind.

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Help in inequalities algebra, please! I am loosing my mind.

by jeffboshine » Sat Feb 18, 2012 10:36 am
I am trying to do this algebraically to understand the concept. I don't want to pick numbers.

Problem 1) A^2<A
Divide both side by positive A, I get A<1
Divide both side by negative A, I get -A<1, or A>1
Obbously Wrong

Problem 2) A/B>C/B
Multiply both side by positive B, I get A>C
Multiply both sides by negative B, I get -A>-C, or A<C
Obviously Wrong

How can I solve these two inequalities algebraically, can someone help with a step by step approach.

Thanks.
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by pemdas » Sat Feb 18, 2012 10:50 am
jeffboshine wrote:I am trying to do this algebraically to understand the concept. I don't want to pick numbers.

Problem 1) A^2<A
Divide both side by positive A, I get A<1
Divide both side by negative A, I get -A<1, or A>1
Obbously Wrong

Problem 2) A/B>C/B
Multiply both side by positive B, I get A>C
Multiply both sides by negative B, I get -A>-C, or A<C
Obviously Wrong

How can I solve these two inequalities algebraically, can someone help with a step by step approach.

Thanks.
the 1st problem is comparatively easier, and you can rewrite the inequality as following
A^2 - A < 0
A(A-1) < 0, A<0 and A>1 (A can't be negative and greater than 1, not possible), therefore look for another case A>0 and A<1. The solution area: 0<A<1

the 2nd problem can be tackled by observing 2 conditions: B<0 or B>0
When B<0, A/B>C/B, A<C
When B>0, A/B>C/B, A>C
you have to know the sign of B to solve the 2nd problem
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by jeffboshine » Sat Feb 18, 2012 11:19 am
Thank you, that clarified a lot.

One more follow up:

(1/7)^4y>(1/7)^8y+14

Here is how I did it:

7^(-4y)>7^(-8y-14)
-4y>-8y-14
-4y>-14
y<14/4

The book says to go from the original statement to
4y<8y+14

I don't understand how they get from the original statement to this one.
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by geno5 » Sat Feb 18, 2012 11:30 am
-4y>-8y-14 from here you need to divide by a - which makes you flip the sign <

therefore

4y<8y+14
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by jeffboshine » Sat Feb 18, 2012 11:36 am
Why is it correct to divide by a negative when you did
and it is wrong to divide by a negative when I did

How do you decide when to do it. I did it later, you did it earlier.
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by geno5 » Sat Feb 18, 2012 11:48 am
I think there may be an error in your calculation I am not sure how you went from
-4y>-8y-14 to
-4y>-14
+8y each side = 4y>-14 therefore y>-14/4

but I chose that point because all have a negative sign and it makes every number positive, seems easy then.
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by jeffboshine » Sat Feb 18, 2012 11:56 am
That was it, a silly mistake, but both your path and my path would lead to the same answer.

Y>-14/4

After looking at this for so long, I am loosing my mind.

Thank you.
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by pemdas » Sat Feb 18, 2012 12:01 pm
jeffboshine wrote:Thank you, that clarified a lot.

One more follow up:

(1/7)^4y>(1/7)^8y+14

Here is how I did it:

7^(-4y)>7^(-8y-14)
-4y>-8y-14
-4y>-14
y<14/4

The book says to go from the original statement to
4y<8y+14

I don't understand how they get from the original statement to this one.
first of all, you make notation mistake!
Not (1/7)^4y>(1/7)^8y+14 BUT (1/7)^4y>(1/7)^(8y+14), I have been wasting a lot of time before figured you must make mistake there in notation

(1/7)^4y>(1/7)^(8y+14)
Never work through powers with fractions, Compare (1/2)^2 > (1/2)^3 doesn't translate into 2>3, but translates into 2<3. 2^-2 > 2^-3 and -2 > -3 or 2 < 3.

Next, (1/7)^4y>(1/7)^(8y+14) must be rewritten as 7^(-4y) > 7^(-8y-14)
-4y > -8y-14, multiply both sides by -1 and flip the sign, 4y<8y+14, -4y<14, again flip the sign, y>-14/4, y>-7/2
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by jeffboshine » Sun Feb 19, 2012 6:18 am
Thank you pemdas, I think I got this inequalities thing figured out.
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