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June 2nd, 2015, 10:27 PM   #41
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$2x-7=5x+6$. We'll solve this equation in steps:
1- Get rid of $2x$ by subtracting $2x$ in both sides: $(2x-7)-2x=(5x+6)-2x$
2-Get rid of brackets and put numbers having x's one next the other: $2x-2x-7=5x-2x+6$
3-You see that $2x-2x=0$ and $-7$ is left alone. In the other side $5x-2x=3x$.
4-From step $3$ the equation rearranges to: $-7=3x+6$.
5-We must cancel out $6$ in this step. To do so, we add the opposite of $6$ in both sides, which is $-6$:
$-7-6=3x+6-6$. Note that $6-6=0$.
6-The equation becomes: $-13=3x$ or $3x=-13$
7-Get rid of $3$. To do so, divide both sides by $3$: $\dfrac{3x}{3}=-\dfrac{13}{3}$.
Note that $\dfrac{3x}{3}=\dfrac{3}{3}x=1*x=x$.
$x=-\dfrac{13}{3}$

Hope that helped.

Last edited by Assayid; June 2nd, 2015 at 10:35 PM.
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June 3rd, 2015, 08:15 AM   #42
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Quote:
Originally Posted by Assayid View Post
$2x-7=5x+6$. We'll solve this equation in steps:
1- Get rid of $2x$ by subtracting $2x$ in both sides: $(2x-7)-2x=(5x+6)-2x$
2-Get rid of brackets and put numbers having x's one next the other: $2x-2x-7=5x-2x+6$
3-You see that $2x-2x=0$ and $-7$ is left alone. In the other side $5x-2x=3x$.
4-From step $3$ the equation rearranges to: $-7=3x+6$.
5-We must cancel out $6$ in this step. To do so, we add the opposite of $6$ in both sides, which is $-6$:
$-7-6=3x+6-6$. Note that $6-6=0$.
6-The equation becomes: $-13=3x$ or $3x=-13$
7-Get rid of $3$. To do so, divide both sides by $3$: $\dfrac{3x}{3}=-\dfrac{13}{3}$.
Note that $\dfrac{3x}{3}=\dfrac{3}{3}x=1*x=x$.
$x=-\dfrac{13}{3}$

Hope that helped.
Why you got banned?
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June 3rd, 2015, 08:36 AM   #43
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Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam.
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June 3rd, 2015, 08:47 AM   #44
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Quote:
Originally Posted by Mohajir View Post
Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam.
Ah I see. WB! Thanks for the help!
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June 3rd, 2015, 10:17 AM   #45
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Well Mathlover, I see you "live nowhere";
perhaps you can move to Banned Camp" too?!
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June 3rd, 2015, 12:39 PM   #46
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Quote:
Originally Posted by Mohajir View Post
Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam.
I'm confuse why and how would you know if the answer is a negative?
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June 3rd, 2015, 12:41 PM   #47
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Quote:
Originally Posted by Denis View Post
Well Mathlover, I see you "live nowhere";
perhaps you can move to Banned Camp" too?!
Why is that?
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June 3rd, 2015, 01:05 PM   #48
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Quote:
Originally Posted by Mathlover44 View Post
I'm confuse why and how would you know if the answer is a negative?
I got a message from mods.
Hey, Mathlover44, don't bother yourself with it anymore. And thanks for your concern.
Please, don't answer me about this. Just keep working math as hard as you can. I want to see you in this forum, understanding and answering high level questions and problems.
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June 3rd, 2015, 01:22 PM   #49
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Please...can someone close this thread...
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