# Inequality: if x+1<0, show that a) 2x-1<0 b) (2x-1)/(x+1) >2

#### mimie

##### New member
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.

#### pka

##### Elite Member
if x+1<0, show that
a) 2x-1<0
\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*}

#### Jomo

##### Elite Member
Or....
x+1<0 given
2x+2<0 multiply by 2
2x - 1<−3 subtract 3
2x - 1< 0 since -3<0

#### Dr.Peterson

##### Elite Member
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.
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.)

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.

#### JeffM

##### Elite Member
Because this is posted in pre-algebra, it is hard to be sure what you know that may be relevant. (Frankly, I am surprised that it is considered a problem appropriate for pre-algebra. It looks like algebra to me.)

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?

#### HallsofIvy

##### Elite Member
If a+ 1< 0 then a+ 1 is negative. Multiplying both sides of an inequality by a negative number changes the direction of the inequality so multiplying both sides of $$\displaystyle \frac{2x- 1}{x+ 1}> 2$$ by x+ 1 gives $$\displaystyle 2x- 1< 2(x+ 1)= 2x+ 2$$. Subtracting 2x from both sides gives $$\displaystyle -1< 2$$, a true statement.

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: $$\displaystyle \frac{2x- 1}{x+ 2}> 2$$.