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 Algebra Pre-Algebra and Basic Algebra Math Forum

 June 2nd, 2015, 10:27 PM #41 Banned Camp   Joined: May 2015 From: Casablanca Posts: 36 Thanks: 7 $2x-7=5x+6$. 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. June 3rd, 2015, 08:15 AM   #42
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 Originally Posted by Assayid $2x-7=5x+6$. 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?  June 3rd, 2015, 08:36 AM #43 Member   Joined: Jun 2015 From: Casablanca Posts: 47 Thanks: 3 Well, I pointed to a link out of general help, and mods banned me seeing it as a commercial spam. June 3rd, 2015, 08:47 AM   #44
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 Originally Posted by Mohajir 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!  June 3rd, 2015, 10:17 AM #45 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Well Mathlover, I see you "live nowhere"; perhaps you can move to Banned Camp" too?! June 3rd, 2015, 12:39 PM   #46
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 Originally Posted by Mohajir 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? June 3rd, 2015, 12:41 PM   #47
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 Originally Posted by Denis Well Mathlover, I see you "live nowhere"; perhaps you can move to Banned Camp" too?!
Why is that? June 3rd, 2015, 01:05 PM   #48
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 Originally Posted by Mathlover44 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. June 3rd, 2015, 01:22 PM #49 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Please...can someone close this thread...  Tags problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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