# solving $0<1-(3x/4)<1$ for $x$

My professor wrote this:

$$0<1-\frac{3x}{4}<1$$

$$-1<-\frac{3x}{4}<0$$

$$4>3x>0$$

$$\frac{4}{3}>x>0$$

Is that correct?

• The first one is closer to be correct, without parenthesis: $0<1-\ 3x/4 <1$. Nov 27 '13 at 0:19
• can you explain why ? and why my is bad ? Nov 27 '13 at 0:20
• The answer should go like $-1<-3x<3$ and so $-1<x<1/3$. Nov 27 '13 at 0:20
• As Berci said, the professor's workings are right if the question is $0<1-3x/4<1$. As for yours, there are mistakes abound. Nov 27 '13 at 0:28
• It is 0<1−(3x/4)<1 With brackets Nov 27 '13 at 0:29

## 1 Answer

Note: This refers to an earlier version of the question, which has now been substantially changed.

Your professor’s step from

$$0<\frac{1-3x}4<1$$

to $$-1<\frac{-3x}4<0\;,$$

is incorrect: $1$ has been subtracted from $0$ and from $1$, but only $\frac14$ has been subtracted from $\frac{1-3x}4$. Had your professor correctly subtracted $\frac14$ from each term, the result would have been $$-\frac14<-\frac{3x}4<\frac34\;.$$

The next step was to multiply everything by $-4$; doing this to the corrected version leaves you with $$1>3x>-3\;,$$ and dividing everything by $3$ then gives you the result, $$\frac13>x>-1\;.$$

The first step of your calculation is clearly wrong: you’ve multiplied the $1$ by $4$ but not the $\frac{1-3x}4$. Judging by your next step, however, I think that you may have meant to multiply everything by $4$ to get $0<1-3x<4$. Then you wanted to subtract $1$, which is fine, but you must subtract it from the $0$ as well as from the $1-3x$ and the $4$, so you should get $-1<-3x<3$. Dividing everything by $-3$ will now give you

$$\frac13>x>-1\;,$$

as before.

• it is not (1-3x)/4 it is 1-(3x/4) Nov 27 '13 at 0:39
• @user2982390: For the revised question your professor’s work is entirely correct. Nov 27 '13 at 0:42
• I don't understand why..or maybe I do, I am very very tired Nov 27 '13 at 0:43
• @user2982390: The steps are: $(1)$ subtract $1$ from each of the three expressions; $(2)$ multiply each expression by $-4$, remembering that this reverses the direction of the inequalities, since $-4$ is negative; $(3)$ divide each expression by $3$, which does not change the direction of the inequalities, since $3$ is positive. Nov 27 '13 at 0:46