YanaSidlinskiy
  • YanaSidlinskiy
The area of a rectangular garden is 54 square meters. The width is 7 meters longer than one-third of the length. Find the length and the width of the garden. Use the formula, are = length * width.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
|dw:1438819879525:dw|
jim_thompson5910
  • jim_thompson5910
The width is 7 meters longer than one-third the length, so we can say \[\Large W = \frac{1}{3}L + 7\]
jim_thompson5910
  • jim_thompson5910
The area of the rectangle is length * width, so \[\Large A = L*W\] \[\Large 54 = L*\left(\frac{1}{3}L + 7\right)\] Do you see how to solve for L from here?

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More answers

YanaSidlinskiy
  • YanaSidlinskiy
Yeah. Would it be \[54=\frac{ 1 }{ 3 }x^2+7x\] From there I need help...
jim_thompson5910
  • jim_thompson5910
what I would do from there is multiply everything by 3 to get rid of the fraction
jim_thompson5910
  • jim_thompson5910
then get everything to one side
YanaSidlinskiy
  • YanaSidlinskiy
So like multiply 1/3x^2(3) and multiply 7x(3)?
jim_thompson5910
  • jim_thompson5910
yeah and 54 as well
YanaSidlinskiy
  • YanaSidlinskiy
Okie, one sec.
YanaSidlinskiy
  • YanaSidlinskiy
Got it. Would it be: \[162=1x^2+21x\]
YanaSidlinskiy
  • YanaSidlinskiy
Or instead I should just put \[x^2\]
jim_thompson5910
  • jim_thompson5910
either works
jim_thompson5910
  • jim_thompson5910
now move that 162 over
YanaSidlinskiy
  • YanaSidlinskiy
Ok. I have a question... Does it matter with what number I multiply the numbers with?
jim_thompson5910
  • jim_thompson5910
you mean like how you multiplied everything by 3 just now?
YanaSidlinskiy
  • YanaSidlinskiy
Yeah.
jim_thompson5910
  • jim_thompson5910
The denominator of the fraction was 3, so multiplying everything by that clears out the denominator (ie eliminates the fraction)
jim_thompson5910
  • jim_thompson5910
if you picked another number that wasn't a multiple of 3, then the denominator or fraction wouldn't go away
YanaSidlinskiy
  • YanaSidlinskiy
Oh ok! Gotcha! Ok. So I got: 1x^2+21x-162=0 Give me a sec. I wanna do it myself and you tell me what I got wrong.
YanaSidlinskiy
  • YanaSidlinskiy
I got: (x+54)(x-3) PLEASE tell me it's right haha! I've had so much trouble with this today.
jim_thompson5910
  • jim_thompson5910
54 times -3 = -162 .... true 54 plus -3 = 21 ... false so that's not the correct factorization
YanaSidlinskiy
  • YanaSidlinskiy
I'm confused and I think this is the stop. So how would I know which number goes where? Or more like why does it matter where I put down the sign with which number?
YanaSidlinskiy
  • YanaSidlinskiy
*spot
jim_thompson5910
  • jim_thompson5910
The quadratic formula may be easier to work with. Did you want to use that?
YanaSidlinskiy
  • YanaSidlinskiy
Yeah, I wanna use that instead... This is really hard for me and I have no idea why...
YanaSidlinskiy
  • YanaSidlinskiy
Ok, let me figure it out using the quadratic formula.
jim_thompson5910
  • jim_thompson5910
1x^2+21x-162 means a = 1 b = 21 c = -162 plug those values into \[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
YanaSidlinskiy
  • YanaSidlinskiy
I'm not exactly sure if this is right, but so far I have: \[x=\frac{ -21\pm \sqrt{207} }{ 2 }\] Is that right?
YanaSidlinskiy
  • YanaSidlinskiy
*-201
jim_thompson5910
  • jim_thompson5910
The 207 portion is incorrect b^2 - 4ac = 21^2 - 4*1*(-162) = 1089
YanaSidlinskiy
  • YanaSidlinskiy
So I multiplied it incorrectly? How?
jim_thompson5910
  • jim_thompson5910
21^2 = 441 4*1*(-162) = -648
jim_thompson5910
  • jim_thompson5910
21^2 - 4*1*(-162) = 441 - (-648) = 1089
YanaSidlinskiy
  • YanaSidlinskiy
Ahh! That's what I was missing! Gotcha! Ok. Hang on:)
YanaSidlinskiy
  • YanaSidlinskiy
So it'd be like \[x=\frac{ -21\pm \sqrt{1089} }{ 2 }\] ?
jim_thompson5910
  • jim_thompson5910
correct
YanaSidlinskiy
  • YanaSidlinskiy
Sorry for making you wait.... I'm kinda slow on thinking right now...
jim_thompson5910
  • jim_thompson5910
You're doing great. Don't worry.
YanaSidlinskiy
  • YanaSidlinskiy
I hope so... So my answer would be x=534?
YanaSidlinskiy
  • YanaSidlinskiy
The answer can never be negative.. I know that...
jim_thompson5910
  • jim_thompson5910
no
YanaSidlinskiy
  • YanaSidlinskiy
Ughgh. Ok, I need some real help. What is it?
jim_thompson5910
  • jim_thompson5910
\[\Large x=\frac{ -21\pm \sqrt{1089} }{ 2 }\] \[\Large x=\frac{ -21+ \sqrt{1089} }{ 2 } \ \text{ or } \ x=\frac{ -21- \sqrt{1089} }{ 2 }\] \[\Large x=\frac{ -21+ 33 }{ 2 } \ \text{ or } \ x=\frac{ -21- 33 }{ 2 }\] \[\Large x=?? \ \text{ or } \ x=??\]
YanaSidlinskiy
  • YanaSidlinskiy
Wow, ok. I divided 1089 by 2 so that's what threw me off... Hang on let me get the final answer..
YanaSidlinskiy
  • YanaSidlinskiy
I got 6, -27.
jim_thompson5910
  • jim_thompson5910
x took the place of the length L so L = 6 or L = -27 but a negative length makes no sense which is why we ignore -27
YanaSidlinskiy
  • YanaSidlinskiy
So, what would be my width? I don't seem to be understanding that part..
jim_thompson5910
  • jim_thompson5910
go back to \[\Large W = \frac{1}{3}L + 7\]
jim_thompson5910
  • jim_thompson5910
plug in L = 6
jim_thompson5910
  • jim_thompson5910
or you can ask yourself "something times 6 gives the area of 54, what is that something?"
YanaSidlinskiy
  • YanaSidlinskiy
Oh yeah. Sorry. I'm like dead tired and didn't get enough sleep lol.. 9
jim_thompson5910
  • jim_thompson5910
yep L = 6 and W = 9
YanaSidlinskiy
  • YanaSidlinskiy
Anyways, THANK YOU SO MUCH!!!!!! I can't appreciate you enough. I'd be sitting behind the computer sending you billions of messages with different thank you's. Thanks again!
jim_thompson5910
  • jim_thompson5910
I'm glad it's making sense now
YanaSidlinskiy
  • YanaSidlinskiy
Yeah, me too. I have another one similar to this, but I'll figure it out myself and let you check it since it's coming to my brain lol. Anyways, just expect a "check message" from me or something...
jim_thompson5910
  • jim_thompson5910
ok sounds like a good plan to me

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