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sar26bentley · Sep 9, 2008, 07:14 PM

I do not understand how to do this Algebra II problem. Solve the open sentence and graph the solution set that is not empty.

5 - a < 3 - 2a and 2 - 3(a-1) < 2 - (a +3)

Unknown008 · Sep 10, 2008, 04:38 AM

You can solve it as a normal equation, but keep in mind that this is not a normal equation. I know, this is a bit difficult at first. Solve the first one.

5 - a < 3 - 2a

a < -2 (you should have this)

and

2 - 3(a-1) < 2 - (a +3)

a > 3 (you should have this)

It is to be noted that when you divide both sides by a negative number, the sign reverses, that is 'less than' becomes 'greater than' and vice versa. Anyway, you can plug in your answer in your original equation to verify your answer. Do the solving first, then we'll go to the graphing.

rebel-2 · Sep 10, 2008, 07:37 PM

As I recall, when you divide or multiply by a negative number the ">/<" signs change accordingly.

jeffk4628 · Jan 22, 2009, 05:27 PM

according to the conditions of the question a<-2 and a>3. Since no value satisfies this, answer is the empty set, but question also says to state solution which is not the empty set. Unsolvable- flawed problem.

Unknown008 · Jan 23, 2009, 10:06 PM

Originally Posted by jeffk4628

according to the conditions of the question a<-2 and a>3. Since no value satisfies this, answer is the empty set, but question also says to state solution which is not the empty set. Unsolvable- flawed problem.

Hey you! I know what I am doing here!! :mad:

Since no value satisfies this

Yes there are, an moreover, there are infinite solutions.

Saying that a<-2 and a>3 is read as "a<-2 or a>3", meaning a cannot be between -2 and 3 but can be any number such as -3, -4, -5, -6, -7... OR 4, 5, 6, 7, 8,.

juhi2011 · Jan 25, 2009, 11:11 AM

find the value of a in both the equations... and that's the answer..