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It looks like extremely simple, but I'm totally confused

5(x + 2) + 2(x-3) < 3(x - 1) + 4x
5x + 10 + 2x - 6 < 3x - 3 + 4x
7x - 4 < 7x - 3
7x - 7x < -3 + 4

0 < 1 !!? <-- How is that?

Why do I lost $x$ variable and it leads to nowhere?! Thanks

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2  
If you had done the computations correctly, it would just tell you the original inequality holds for every $x$. But you didn't... – David Mitra Dec 12 '12 at 17:36
How can it hold for all $x$? Setting $x=0$ gives $4<-3$ which is clearly false. – copper.hat Dec 12 '12 at 17:42
2  
There is a mistake on line 3 of the computation. The $-4$ should be $+4$. – copper.hat Dec 12 '12 at 17:45

4 Answers

up vote 4 down vote accepted

If you simplify, you obtain the inequality $7x+4 < 7x-3$ which is the same as $4<-3$. Since this is never true, the inequality is false, regardless of the value of $x$.

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up vote 3 down vote

You are correct. The conclusion you came to means that for all $x$ the inequality you stated is true

EDIT: As De Vito pointed out, if you arrive correctly at a false statement this necessarily means that the assumption is false. In other words if the implication $P\Rightarrow Q$ is true and $Q$ is false then $P$ is also false.

But, if you arrive correctly at a true statement, this doesn't necessarily mean that the assumption is true. For that you will need $P\Leftrightarrow Q$ and not simply $P\Rightarrow Q$. In our case, you have shown $5(x + 2) + 2(x-3) < 3(x - 1) + 4x\Leftrightarrow 4<-3$

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$0<1$ is true. You might want to fix that and you receive an upvote. – CBenni Dec 12 '12 at 17:35
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No, it is true for all $x$. – Ross Millikan Dec 12 '12 at 17:36
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I corrected it. – Nameless Dec 12 '12 at 17:36
Careful: Proving a true statement from an assumption often does NOT mean the assumption is true. (In this case it does, but more should be said). – Jason DeVito Dec 12 '12 at 17:37
The statement is never true. – copper.hat Dec 12 '12 at 17:44
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up vote 4 down vote

Getting a result like $0<1$ means that the claim is true for all x. In other words, for any $x$ you input, the inequation holds true. (For all of your steps taken were equivalent to their predecessor, and as such did not change the truth/false value of the inequation. In some cases, mathematicians like to denote that by adding "$\Leftrightarrow$" at the start of each line.)

And there is a error in your computation aswell. I know no (reasonable) definition of $\mathbb{Z}$ such that $10-6=-4$

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Careful: Proving a true statement from an assumption often does NOT mean the assumption is true. (In this case it does, but more should be said). – Jason DeVito Dec 12 '12 at 17:38
Mmh. Do you mean I should mention that all of the steps taken are equivalent to their predecessor? – CBenni Dec 12 '12 at 17:41
@JasonDeVito: It does when the steps are reversible, which they are here (and correct, too, which needs some work) – Ross Millikan Dec 12 '12 at 17:42
Yup, that would fix it. – Jason DeVito Dec 12 '12 at 17:42
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I upvoted 3 answers cause they are informative enough to me and yours as well +1 – metal_fan Dec 12 '12 at 18:01
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up vote 0 down vote

It's like $a_{1}y+a_{1}y+……a{n}y+C<b_{1}y+b_{2}y+……b{m}y+D$

if$\sum a_{i}=\sum b_{j}$ the unkown y will certainly disappear. You can think like this.

Let A=1+2+4

Let B=3+5+(-1)

Then A=B, A and B are coefficient of y for the left and right.

Then y disappear.

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