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hunterhoff
Group Title
Create your own twostep equation with All Real Numbers as the solution. Show the steps to solving the equation to prove the solution All Real Numbers.
My answer is:
2(x + 3) = 2(x + 1)  4
2x6 =2x24
2x6=2x6
2x=2x
x=2x/2
Is this right?
 one year ago
 one year ago
hunterhoff Group Title
Create your own twostep equation with All Real Numbers as the solution. Show the steps to solving the equation to prove the solution All Real Numbers. My answer is: 2(x + 3) = 2(x + 1)  4 2x6 =2x24 2x6=2x6 2x=2x x=2x/2 Is this right?
 one year ago
 one year ago

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UmmmHELP Group TitleBest ResponseYou've already chosen the best response.0
I think...
 one year ago

UmmmHELP Group TitleBest ResponseYou've already chosen the best response.0
are there other options or just that one?
 one year ago

hunterhoff Group TitleBest ResponseYou've already chosen the best response.0
There were no options, I had to come up with one
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Well, this problem is a case in which \(\ \Huge x=All Real Numbers \).
 one year ago

hunterhoff Group TitleBest ResponseYou've already chosen the best response.0
Lol so is it correct?
 one year ago

UmmmHELP Group TitleBest ResponseYou've already chosen the best response.0
Yeah im not sure...sorry!
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
One moment...
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
So, what you have is mostly correct. dw:1351659038210:dw Because we have the same expressions on both sides of the equation sign, x could be any real number, and the equation would render true. Does that make sense?
 one year ago

hunterhoff Group TitleBest ResponseYou've already chosen the best response.0
Umm I think so
 one year ago

hunterhoff Group TitleBest ResponseYou've already chosen the best response.0
Would it meet my teachers approval?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
The correct answer then, for this problem, would be "All Real Numbers". Additionally, your question states prove that the solution to the eqn is all real numbers.
 one year ago

hunterhoff Group TitleBest ResponseYou've already chosen the best response.0
Oh ok, thanks! :)
 one year ago
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