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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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    There is a mistake on line 3 of the computation. The $-4$ should be $+4$.2012-12-12

4 Answers 4

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It's like $a_{1}y+a_{1}y+……a{n}y+C

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.

4

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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    Someone just downvoted all answers. Felt no need to explain however.2012-12-12
4

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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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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    @copper.hat Corrected2012-12-12