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.

- YanaSidlinskiy

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- jim_thompson5910

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- 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

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

Yeah. Would it be \[54=\frac{ 1 }{ 3 }x^2+7x\]
From there I need help...

- jim_thompson5910

what I would do from there is multiply everything by 3 to get rid of the fraction

- jim_thompson5910

then get everything to one side

- YanaSidlinskiy

So like multiply 1/3x^2(3) and multiply 7x(3)?

- jim_thompson5910

yeah and 54 as well

- YanaSidlinskiy

Okie, one sec.

- YanaSidlinskiy

Got it. Would it be:
\[162=1x^2+21x\]

- YanaSidlinskiy

Or instead I should just put \[x^2\]

- jim_thompson5910

either works

- jim_thompson5910

now move that 162 over

- YanaSidlinskiy

Ok. I have a question... Does it matter with what number I multiply the numbers with?

- jim_thompson5910

you mean like how you multiplied everything by 3 just now?

- YanaSidlinskiy

Yeah.

- jim_thompson5910

The denominator of the fraction was 3, so multiplying everything by that clears out the denominator (ie eliminates the fraction)

- jim_thompson5910

if you picked another number that wasn't a multiple of 3, then the denominator or fraction wouldn't go away

- 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

I got:
(x+54)(x-3)
PLEASE tell me it's right haha! I've had so much trouble with this today.

- jim_thompson5910

54 times -3 = -162 .... true
54 plus -3 = 21 ... false
so that's not the correct factorization

- 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

*spot

- jim_thompson5910

The quadratic formula may be easier to work with. Did you want to use that?

- YanaSidlinskiy

Yeah, I wanna use that instead... This is really hard for me and I have no idea why...

- YanaSidlinskiy

Ok, let me figure it out using the quadratic formula.

- 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

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

*-201

- jim_thompson5910

The 207 portion is incorrect
b^2 - 4ac = 21^2 - 4*1*(-162) = 1089

- YanaSidlinskiy

So I multiplied it incorrectly? How?

- jim_thompson5910

21^2 = 441
4*1*(-162) = -648

- jim_thompson5910

21^2 - 4*1*(-162) = 441 - (-648) = 1089

- YanaSidlinskiy

Ahh! That's what I was missing! Gotcha! Ok. Hang on:)

- YanaSidlinskiy

So it'd be like \[x=\frac{ -21\pm \sqrt{1089} }{ 2 }\]
?

- jim_thompson5910

correct

- YanaSidlinskiy

Sorry for making you wait.... I'm kinda slow on thinking right now...

- jim_thompson5910

You're doing great. Don't worry.

- YanaSidlinskiy

I hope so... So my answer would be x=534?

- YanaSidlinskiy

The answer can never be negative.. I know that...

- jim_thompson5910

no

- YanaSidlinskiy

Ughgh. Ok, I need some real help. What is it?

- 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

Wow, ok. I divided 1089 by 2 so that's what threw me off... Hang on let me get the final answer..

- YanaSidlinskiy

I got 6, -27.

- 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

So, what would be my width? I don't seem to be understanding that part..

- jim_thompson5910

go back to \[\Large W = \frac{1}{3}L + 7\]

- jim_thompson5910

plug in L = 6

- jim_thompson5910

or you can ask yourself "something times 6 gives the area of 54, what is that something?"

- YanaSidlinskiy

Oh yeah. Sorry. I'm like dead tired and didn't get enough sleep lol.. 9

- jim_thompson5910

yep L = 6 and W = 9

- 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

I'm glad it's making sense now

- 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

ok sounds like a good plan to me

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