System equations

[Loki123]

I honestly have no idea how to solve this, but I tried. Any advice?

 

[Steven G]

This is not a system of equations! An equation has an equal sign. This is a system of inequalities.

 

[Steven G]

I would carefully divide by a-1 and by a+1.

 

[Loki123]

I would carefully divide by a-1 and by a+1.

Divide what exactly?

 

[topsquark]

I honestly have no idea how to solve this, but I tried. Any advice? View attachment 32521View attachment 32522

What Steven G means is
[imath]a(a - 1)x \geq a - 1[/imath]

Divide both sides by a - 1. Be careful with the cases where a - 1 > 0 and a - 1 < 0. If, for example, a - 1 > 0 can a + 1 < 0?

-Dan

 

[Loki123]

What Steven G means is
[imath]a(a - 1)x \geq a - 1[/imath]

Divide both sides by a - 1. Be careful with the cases where a - 1 > 0 and a - 1 < 0. If, for example, a - 1 > 0 can a + 1 < 0?

-Dan

what about a-1=0

 

[BigBeachBanana]

what about a-1=0

Can you divide 0?

 

[Steven G]

Divide what exactly?

Isn't it obvious? Divide the inequality by a-1. Divide the other inequality by a+1.

 

[Loki123]

Can you divide 0?

what i mean is that by diving by a-1 i am introducing a condition that a-1 can't be 0, but what if a is 1?

 

[Loki123]

Isn't it obvious? Divide the inequality by a-1. Divide the other inequality by a+1.

can you explain why? what is the goal?

 

[Steven G]

Why is it obvious that I wanted you to divide the inequality by a-1. What else would you divide by a-1??!!!
Dividing the appropriate inequality would make the inequality much cleaner looking. Do you see that?

If you had 7x^2 < 14x + 7, wouldn't you divide by 7 to simplify it to x^2<2x+1.
Then x^2-2x+1 = (x-1)^2<0. So the solution is the empty set.

 

[Steven G]

what i mean is that by diving by a-1 i am introducing a condition that a-1 can't be 0, but what if a is 1?

Do that case separately. I did suggest that you divide carefully.

 

[Loki123]

Do that case separately. I did suggest that you divide carefully.

i usually avoid diving because it can eliminate a certain solution
for example if i were to divide by (x-1) some equation i am saying that x cannot be 1 because than i would be diving by 0 and I have actually come across cases where x would turn out to be 1. So I not sure I get it and I don't want to memorize this problem...

 

[Steven G]

You can solve this by 3 cases.
Case 1: a-1<0
Case 2: a-1>0
Case 3: a-1 = 0.

Try solving the 1st inequality for each case.

 

[Loki123]

You can solve this by 3 cases.
Case 1: a-1<0
Case 2: a-1>0
Case 3: a-1 = 0.

Try solving the 1st inequality for each case.

It would appear only third statement is correct. Do you agree?

 

[Steven G]

Under I, you wrote if a>1, then a>-1. Is that what you meant to write?

 

[Steven G]

Can you please solve this simple inequality: -2x<12

 

[Loki123]

Under I, you wrote if a>1, then a>-1. Is that what you meant to write?

Yes. If I say a>1, a has to also >-1

 

[Loki123]

Can you please solve this simple inequality: -2x<12

x>-6

 

[Steven G]

x>-6

It seemed that I saw somewhere in your work you failed to interchange the inequality sign.