- #1

inverse

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Hello!

a)[itex]\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

b)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}[/itex]

c)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

Can you tell me how I can resolve it?

In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2,

but when I pass 2x-1 to multipliy with 1 "(3<1·2x-1)" Why I have not change "<" by ">" The denominator is negative, when is negative I thought I had change it.

The solution set is [itex](-\infty,\frac{1}{2})\cup(2,+\infty)[/itex]

for x> 2 not to have to change the sign to move the denominator to multiply by 1.

Thank you very much for your help :)

a)[itex]\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

b)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}[/itex]

c)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

Can you tell me how I can resolve it?

In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2,

but when I pass 2x-1 to multipliy with 1 "(3<1·2x-1)" Why I have not change "<" by ">" The denominator is negative, when is negative I thought I had change it.

The solution set is [itex](-\infty,\frac{1}{2})\cup(2,+\infty)[/itex]

for x> 2 not to have to change the sign to move the denominator to multiply by 1.

Thank you very much for your help :)

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