anonymous
  • anonymous
algebra
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
y=5/3x+8
whpalmer4
  • whpalmer4
Okay, \[y+2 = \frac{5}{3}(x+6)\]You want \[?x + ?y = ?\] Any ideas how to start?

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anonymous
  • anonymous
x=3y5−245
anonymous
  • anonymous
y=8+5x/3
whpalmer4
  • whpalmer4
Why don't we get rid of that fraction first? Multiply everything by the denominator of the fraction: \[3*y + 3*2 = 3*\frac{5}{3}(x+6)\]\[3y + 6= 5(x+6)\] Can you rearrange that so that the \(x\) and \(y\) terms are on the left, and anything else on the right?
anonymous
  • anonymous
y=5/3x+8
whpalmer4
  • whpalmer4
Why do you keep typing that stuff?
whpalmer4
  • whpalmer4
Do you know what the distributive property is?
anonymous
  • anonymous
im lost
whpalmer4
  • whpalmer4
Do you know what the distributive property is? \[a(b+c) = a*b + a*c\] it's how you get rid of parentheses like we have in \[3y+6=5(x+6)\] Can you use it to get rid of the parentheses, please?
anonymous
  • anonymous
y=13 x
whpalmer4
  • whpalmer4
Come on...you had \[3y+6 = 5(x+6)\]and you are telling me that becomes \(y = 13x\) I want you to do just one simple step: use the distributive property to remove the parentheses. Do not simplify, do not collect like terms, leave everything there exactly as it is if you do only that one step.
anonymous
  • anonymous
3y+6=5x+6
whpalmer4
  • whpalmer4
Well, closer, but no. \[5(x+6) = 5*x + 5*6 = 5x + 30\] So, \[3y + 6 = 5(x+6)\]\[3y+6 = 5*x + 5*6\]\[3y+6 = 5x +30\] Now, can you rearrange that so that the terms with \(x\) and \(y\) in them are on the left-hand side, and the terms with only numbers are on the right-hand side?
anonymous
  • anonymous
anonymous
  • anonymous
?????
whpalmer4
  • whpalmer4
Please do what I ask. We'll save a lot of time if you do that. \[3y+6 = 5x+30\]Rearrange that so that you have an equation with the terms containing \(x\) or \(y\) on the left side of the = and any terms not containing one of them on the right side of the =.
whpalmer4
  • whpalmer4
Let's look at the first term: \(3y\) Do we want that on the left-hand side or the right-hand side?
anonymous
  • anonymous
+30 its wrong
MD152727
  • MD152727
MD152727
  • MD152727
please medal as I spent lots of time in this
whpalmer4
  • whpalmer4
@MD152727 that is NOT what the problem asks you to do.
whpalmer4
  • whpalmer4
problem wants the equation in standard form: \[Ax + By = C\]
MD152727
  • MD152727
:? *facepalm*
whpalmer4
  • whpalmer4
Yeah, I know the feeling :-)
MD152727
  • MD152727
My gosh!
whpalmer4
  • whpalmer4
Helpful tip that has saved me countless times over the years: when you think you have an answer, check your work. When you have checked and are sure that you didn't make a stupid mistake somewhere, re-read the problem and make sure you have answered the question asked by the problem, and not some other question.
MD152727
  • MD152727
That is what I understood! I am an 11 years old in 9th grade
anonymous
  • anonymous
I'm still not get the problem solve
MD152727
  • MD152727
ok, sorry ignore me, i annoyed you two
anonymous
  • anonymous
??
whpalmer4
  • whpalmer4
Perhaps if you did as I requested, @EKKERKING, we could be done... I will ask again: the term \(3y\), do we want to keep it on the left side or move it to the right side?
anonymous
  • anonymous
move it
whpalmer4
  • whpalmer4
Let me remind you of my instructions: Rearrange that so that you have an equation with the terms containing x or y on the left side of the = and any terms not containing one of them on the right side of the =. Do we want to move \(3y\) to the right side, or keep it on the left side?
anonymous
  • anonymous
leave it like that
whpalmer4
  • whpalmer4
Excellent! That's correct. Here's what we have so far, just so we can see it: \[3y + 6=5x+30\]We decided that the \(3y\) stays on the left. How about the \(+6\), keep on the left, or move to the right?
anonymous
  • anonymous
move it to the right
whpalmer4
  • whpalmer4
Very good. How do we do that?
anonymous
  • anonymous
3y+5x=6+30
whpalmer4
  • whpalmer4
No, I asked "how?" \[3y+6 = 5x+30\]If we to move that \(+6\) out of the left-hand side, we must subtract it from both sides. \[3y+6 -6 = 5x + 30 -6\]\[3y = 5x + 24\] Any question about that?
anonymous
  • anonymous
bro I'm only missing 1 number
anonymous
  • anonymous
whpalmer4
  • whpalmer4
Do you understand what I did there?
anonymous
  • anonymous
no
whpalmer4
  • whpalmer4
What grade are you in?
anonymous
  • anonymous
7
anonymous
  • anonymous
help me with that number please
whpalmer4
  • whpalmer4
Just curious. \[3y + 6 = 5x + 30\]You can do anything you want, pretty much, as long as you do it to both sides of the equation. We have a \(+6\) on the left side, but we don't want it there. To get rid of it, we subtract \(6\) because \(+6 - 6 = 0\). But to subtract it from the left side, we also need to subtract it from the right side or we won't have a true equation any longer. \[3y + 6 - 6 = 5x + 30 -6\]\[3y +\cancel{6 - 6} = 5x + 30 - 6\]\[3y = 5x + 30 - 6\]But \(30-6 = 24\) so we can simplify the right hand side as well: \[3y = 5x +24\]
whpalmer4
  • whpalmer4
Now, how about that \(5x\) term: keep it where it is, or move it to the left side?
anonymous
  • anonymous
it was -24 thanks buddy
whpalmer4
  • whpalmer4
Good luck!

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