- Thread starter mimie
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\(\displaystyle \begin{align*}x+1&<0\text{ given} \\2x+2&<0\text{ multiply by }2\\2x&<-2\text{ subtract }2\\2x&<-2<1\\2x-1&<0\text{ subtract }1\text{ substract } \end{align*}\)if x+1<0, show that

a) 2x-1<0

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Another way: First solve the given inequality for x, then multiply the resulting inequality by 2 and subtract 1 in order to get 2x-1 on the left-hand side. Then observe that if 2x-1 < -3, it is also less than 0, so (a) is implied by the original. (It is not stated that the inequalities should be equivalent, only that one implies the other.)if x+1<0, show that

a) 2x-1<0

b) (2x-1)/(x+1) >2

How to do this question , pls help me. Thanks.

Then see what you can do for (b). I can think of several ways to do it; one is to rewrite the LHS by long division as 2 plus a fraction, and think about the sign of that fraction.

With reference to b,

\(\displaystyle \text {If } a < 0 \text { and } b < 0,\)

\(\displaystyle \text {is } a \times b < 0, a \times b = 0, \text { or } a \times b > 0?\)

How is the answer to the preceding question relevant to your problem?

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For a more "rigorous" proof, go the other way: \(\displaystyle -1< 2\) is obviously true. Add 2x to both sides to get \(\displaystyle 2x- 1< 2x+ 2= 2(x+ 1)\). Divide both sides by the negative number x+ 1: \(\displaystyle \frac{2x- 1}{x+ 2}> 2\).