Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

solve for the inequality 3x-2/x-4 greater than or equal to 0..help please.

OpenStudy (mathstudent55):

Do you mean \[\frac{ 3x - 2 }{ x - 4 } \ge 0\]

OpenStudy (anonymous):

yes..

OpenStudy (anonymous):

whoa hold the phone

OpenStudy (anonymous):

there is a big mistake here

OpenStudy (anonymous):

you cannot multiply by any expression with a variable, because you don't know it it is positive or negative lets start again

OpenStudy (anonymous):

\[\frac{ 3x - 2 }{ x - 4 } \ge 0\] numerator changes sign at \(\frac{2}{3}\) denominator changes sign at \(4\)

OpenStudy (anonymous):

break up the real line in to three intervals \[(-\infty,\frac{2}{3}),(\frac{2}{3},4),(4,\infty)\]

OpenStudy (anonymous):

your expression will change signs at each of these intervals. it will be positive, then negative, then positive

OpenStudy (anonymous):

you can check the sign by replacing \(x\) by any number you choose in any interval for example if \(x=0\) then you get \(\frac{-2}{-4}=\frac{1}{2}\) which means it is positive on the interval containing 0, namely \[(-\infty,\frac{2}{3})\]

OpenStudy (anonymous):

so it is positive on \((-\infty, \frac{2}{3})\) negative on \((\frac{2}{3},4)\) and then positive again on \((4,\infty)\)

OpenStudy (anonymous):

you want positive, or more correctly "greater than or equal to zero" so pick \[(-\infty, \frac{2}{3}]\cup (4,\infty)\]

OpenStudy (anonymous):

4 is "open" because the denominator cannot be zero, \(\frac{2}{3}\) is closed because that is when the expression is equal to zero

OpenStudy (anonymous):

be careful with these, it is tempting to multiply by the denominator to clear the fractions, but you cannot do that. again the reason is that you do not know if it is positive or negative, so you do not know whether to change the inequality or not

OpenStudy (mathstudent55):

Thanks satellite73. You are correct. I solved it like an equation instead of an inequality. I've removed my incorrect answer, so it'll be less confusing. I just left my first question on top to clear up what the problem reads like.

OpenStudy (anonymous):

okay..thanks satellite73. && also thank you mathstudent55 for helping also. ya are the best (:

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!