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Mathematics 8 Online
OpenStudy (anonymous):

The area of a rectangular piece of land is 280 square meters. If the length of the land was 5 meters less and the width was 1 meter more, the shape of the land would be a square. Part A: Write an equation to find the width (x) of the land. Show the steps of your work. (5 points) Part B: What is the width of the land in meters? Show the steps of your work. (5 points)

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

Help please

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

No

OpenStudy (anonymous):

can you post the links over here

OpenStudy (anonymous):

I already have.

OpenStudy (anonymous):

but its just one, can you post a few

OpenStudy (anonymous):

@ganeshie8 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

OpenStudy (anonymous):

@ganeshie8 may you please help me, I've been stuck on this for so long

OpenStudy (anonymous):

@ganeshie8 I will give you a medal

ganeshie8 (ganeshie8):

say the width of original land is \(x\) and length is \(y\)

OpenStudy (anonymous):

ok

ganeshie8 (ganeshie8):

`If the length of the land was 5 meters less and the width was 1 meter more, the shape of the land would be a square.`

ganeshie8 (ganeshie8):

what do you know about sides of a square ?

OpenStudy (anonymous):

^2

OpenStudy (anonymous):

since there are two sides

ganeshie8 (ganeshie8):

we use the fact that the sides of a square are equal

OpenStudy (anonymous):

ok

ganeshie8 (ganeshie8):

\(\large y-5 = x+1 \tag{1} \)

OpenStudy (anonymous):

That's the equation?

ganeshie8 (ganeshie8):

since the area of original land is 280 : \[\large xy = 280\tag{2}\]

OpenStudy (anonymous):

ok I am understanding so far

ganeshie8 (ganeshie8):

isolate y from first equation and substitute it in 2nd equation

ganeshie8 (ganeshie8):

from first equation you get : \(\large y = x + 6\) right ?

OpenStudy (anonymous):

Yes

ganeshie8 (ganeshie8):

plug that in second equation, you will get a quadratic : \[\large xy = 280\\x(x+6) = 280\\x^2+6x-280=0\]

ganeshie8 (ganeshie8):

thats the final equation for part A

ganeshie8 (ganeshie8):

solve that equation for part B

OpenStudy (anonymous):

SO x^2+6x-280?

ganeshie8 (ganeshie8):

x^2+6x-280 = 0

ganeshie8 (ganeshie8):

an equation must have an "=" symbol somewhere

OpenStudy (anonymous):

WHat do I solve for

ganeshie8 (ganeshie8):

x

OpenStudy (anonymous):

I got 14 and -20

ganeshie8 (ganeshie8):

can the width be negative ?

OpenStudy (anonymous):

no

ganeshie8 (ganeshie8):

so throw away -20

OpenStudy (anonymous):

and 14 is correct?

ganeshie8 (ganeshie8):

width = x = 14

OpenStudy (anonymous):

Are you sure?

ganeshie8 (ganeshie8):

if you're sure, im also sure

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