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
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
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.
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 $10-6=-4$
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 $x$.
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 $5(x + 2) + 2(x-3) < 3(x - 1) + 4x\Leftrightarrow 4<-3$