keana
  • keana
help! (picture in comments)
Mathematics
katieb
  • katieb
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keana
  • keana
zepdrix
  • zepdrix
Ok let's see what we know so far...\[\large\rm \color{green}{P}=2\color{royalblue}{L}+2\color{orangered}{W}\] They told us that the \(\large\rm \color{green}{\text{perimeter is 212}}\). So let's plug that in.\[\large\rm \color{green}{212}=2\color{royalblue}{L}+2\color{orangered}{W}\]
zepdrix
  • zepdrix
They told us that the \(\large\rm \color{royalblue}{\text{Length is two more than the Width}}\). We can write that relationship like this: \(\large\rm \color{royalblue}{L=W+2}\)

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keana
  • keana
yes
zepdrix
  • zepdrix
\[\large\rm \color{green}{212}=2\color{royalblue}{(L)}+2\color{orangered}{W}\]We plug this information into our equation, replacing L.
keana
  • keana
right
zepdrix
  • zepdrix
\[\large\rm \color{green}{212}=2\color{royalblue}{(W+2)}+2\color{orangered}{W}\]
zepdrix
  • zepdrix
And from there, you have an equation involving only ONE VARIABLE! :) So you can proceed to solve for W.
zepdrix
  • zepdrix
What do you think? :O
keana
  • keana
lemme work it out
keana
  • keana
212=4w+4?
zepdrix
  • zepdrix
So you ditributed and then combined like-terms? Ya that's a good start! :)
keana
  • keana
52=w?
zepdrix
  • zepdrix
Looks good! And then don't forget about the fact that the \(\large\rm \color{royalblue}{\text{Length is two more than the Width}}\).
keana
  • keana
so 54?
zepdrix
  • zepdrix
yay good job \c:/
keana
  • keana
thanks!

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