The width of a rectangle is 6 in. less than its length. The perimeter is 68 in.
What is the width of the rectangle?
in.

- gabbyalicorn

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- gabbyalicorn

Hi pooja.

- pooja195

Uh i think the answer is in the question....if im right

- anonymous

divide it in two first

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## More answers

- Jhannybean

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- Jhannybean

Omg I forgot the shapes have the "fill" option.... anyway.

- anonymous

yeah

- gabbyalicorn

When you say divide two first, do you mean 68 ÷ 2 and 6÷ 2

- anonymous

no 68 by 2

- anonymous

then subtract 6 from that

- gabbyalicorn

okay, that is 34...

- Jhannybean

the length of the rectangle is given, but they say the `width` `is (equal to)` the `length - 6 in`You can rewrite this as \(\sf w = L-6\)

- Jhannybean

So the Perimeter of a rectangle is represented as: \[\sf P = 2L +2W\]

- Jhannybean

Since we're given the perimeter, and that is = 68 in. we can substitute that in to our formula. \[\sf 68 = 2L +2W\]

- Jhannybean

But we also know something else, \(\sf W=L-6\)

- gabbyalicorn

68 - 6?

- Jhannybean

So we can substitute this in to the formula for the perimeter. \[\sf 68 = 2L+2(L-6)\]

- Jhannybean

Are you following, @gabbyalicorn ?

- gabbyalicorn

Um, somewhat... I don't really get it but i'm following the steps if that's what you mean. :)

- Jhannybean

Which part are you confused with? Let's clarify that before moving on.

- gabbyalicorn

is L the variable we are trying to find

- gabbyalicorn

the number for...

- Jhannybean

We are trying to find W (width). The problem indiscreetly gives us the equation for the width by saying that the width IS 6 LESS THAN the length. We can write an equation for that. \(\sf w = L-6\)

- gabbyalicorn

but what is L

- Jhannybean

But we can't automatically find the width, \(\sf w\) can we? We need to find the length, \(\sf L\) first.

- Jhannybean

That is where we use the equation for the perimeter of the rectangle. \(\sf P = 2L +2w\)

- Jhannybean

Do you see how this works?

- gabbyalicorn

I think I'm getting it...

- Jhannybean

Okay, so we have \[\sf 68 = 2L+2(L-6)\] We need to first distribute (meaning multiply) 2 to each term inside the parenthesis. \[2 \cdot L =~?\]\[2\cdot (-6) =~?\]

- gabbyalicorn

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- Jhannybean

Good

- gabbyalicorn

:)

- Jhannybean

\[\sf 68 = 2L +2L-12\]We add +12 to both sides of the equation.

- gabbyalicorn

why - 12 I thought you said add

- Jhannybean

because we want to isolate the L's to one side of the equation, we need to eliminate the 12 from that side. In order to eliminate it we need to add +12 to it. \[\sf +12-12 = 0\]So, \(\sf 68 + 12 =~?\)

- gabbyalicorn

80

- Jhannybean

So now we have \[\sf 80 = 2L + 2L\]\[\sf 2L + 2L =~?\]

- gabbyalicorn

4L ?

- Jhannybean

Good.

- gabbyalicorn

:}

- Jhannybean

\[\sf 80 = 4L\]Now we want to isolate the right side of the equation so we have JUST L. This is when we divide BOTH sides of the equation by 4. \[\sf \frac{80}{4}=~?\]

- gabbyalicorn

20

- Jhannybean

ALRIGHT! So now we have figured out \(\sf L\). \[\sf L=20\]Now we can go back to our equation for width and plug in L. \[\sfW=L-6\]So therefore what does W =?

- Jhannybean

Sorry, typo. \(\sf W=L-6\)*

- gabbyalicorn

20 - 6 = 14?

- Jhannybean

That's correct :)

- gabbyalicorn

Yes. :)

- Jhannybean

Great! You worked it out yourself :)

- gabbyalicorn

:D Thank you for helping me and not giving up on me! I reallu appreciate it. :)

- Jhannybean

No problem ^^

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