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anonymous
 4 years ago
what is 7yy=1
anonymous
 4 years ago
what is 7yy=1

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theEric
 4 years ago
Best ResponseYou've already chosen the best response.0It's an equation; would you like to know what exact number y is?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yy=17 y^2=8 (y^2=8) \[\sqrt{y}=\sqrt{8}\] y= \[2\sqrt{2}\]

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0was this \[7yy = 1\]or\[7*(y)*(y)=1\]\[7y*(y) = 1\] ? I have to know before I go on!

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0@Veneria (y)  y, or (y) + (y), is actually 2(y). It can also be written as 2(y) or 2y.

theEric
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know if you wanted to know what y is or not, but I'll help, just in case. You start with\[7yy=1\] and now I'll add y to both sides, twice. I hope you're familiar with the concept of adding something to both sides. The overall point is that you change both sides in the same way  so that if they are equal before (like they are) then they are still equal after. That way you know your equation is correct. Variable values don't change, so they are the same in the new equation as well. I go from what I had before, to:\[7yy+y+y=(1)+y+y\]\[=\] and y+y is 2 y's added together. That means y+y = 2*y. That's another way of writing it :) So I'll do that, make it happen.\[7=(1)+2y\] And subtract (1) from both sides. I could also add 1, because that makes more sense, but I won't. This way you see more algebra; I'm subtracting a negative number. It's hard to get your head around, but it's actually adding that number. \[7(1)=(1)(1)+2y\]\[=\]\[7+(+1)=(1)+(+1)+2y\]\[=\]\[7+1=(1)+1+2y\]\[=\]\[8=2y\] Now divide both sides by 2..... I have to go. Good luck!
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