Width - Driveway

[Guest]

I don't know. Never mind.

[Guest]

Are you just wanting the answer for this problem OR do you want to know how to solve it yourself.....

I had asked you this earlier.....
We had a piece of wood 5 foot long when we cut a bit "X" off it we were left with 3 feet.......How much did we cut off.?
Answer..... We cut off "X" feet. The question told you that, all you had to do was read it....but the piece of wood was 5 feet to start with....and we were left with 3 feet when we cut "X" off the 5 feet piece. So that means the bit we were left with equals 3 feet. So if we started with 5 feet and cut off "X" feet, that is the same as saying we started with 5 feet and took away "X" feet...Hows much would be left....The question says we were left with 3 feet.....So 5 feet take away "X" feet equals 3 feet left.
So (5 - X) = 3....rearranging gives... X = (5 - 3)......So X = 2 feet.......so we cut off 2 feet.
That is same as saying we had 5 feet and cut off 2 feet so we are left with 3 feet.

Now do the same when looking at the front of the house......
When we are standing on the road looking at the front of the house and we know the length of the side of the plot and there is a driveway "W" on each side and the front width of the house in between......What is the width of the house?

We know the length of the whole front side of the plot is 100 feet. And we said we would let the driveway width be "W". There is a driveway at each side of the house and the house is in the middle.
So the whole front of the plot is made up of "W driveway" + "front of house" + "W driveway" and we know that the total of this is 100 feet. So in algebra we write this as ( W + Front + W) = 100....OR rearranging....(Front + 2W )= 100.......
OR...rearranging again...... Front = (100 - 2W). If we call the front the width of the house then.... Width = (100 - 2W) feet.
To be able to work out the area of the house we need to work out the front to back dimension as well, call this the length.

Looking sideways at the house, the side of the plot is made up of "W driveway at back" + "front to back or Length" = 100....
So in algebra (W + Length) = 100......OR Length = (100 - W)

Now all you have to do is multiply the length by the width and put it equal to 5000 sq feet as an equation and solve......

[Guest]

(100 - 2W) (100 - W)

Using FOIL:

10000 - 100W - 200W - 100W

10000 - 400W

??

[Guest]

I don't undestand how you were not able to understand subtracting "X" for a piece of wood an few posts ago....and now you have heard about "FOIL" for multiplying out expressions.....Yet you seem to not know to make it equal 5000 to make it into an equation as has been explained several times in these posts..........

Below is complete solution so hopefully you will have no bother understanding it.................

A house is to be built on a square plot of land.
The house will face the street, with a driveway around the other 3 sides.
The owner wishes the house to cover the same amount of land as the driveway. Determine width of driveway, with the plot being 100 feet each side.

The area of the plot is [tex]100 \times 100 = 10000[/tex] sq.feet.
The area of the driveway is half the plot [tex]= 5000[/tex] sq.feet So this is also the area of the house.
The driveway is "around the other 3 sides" so that means the house is at the front edge (in the middle) of the plot facing the street and a driveway is along each side and along the back of the house.
All of the driveway is the same width.
Let the width of the driveway be [tex]= "w"[/tex] feet.

To solve we need to write an expression for the driveway or the house in terms of [tex]"w"[/tex] and put it equal to [tex]5000[/tex].
The house is all one rectangle and the driveway is made up of 3 rectangles so the expression for the house should be more straightforward.

The width of the house is [tex](100 - 2w)[/tex] and the front to back dimension is [tex](100 - w)[/tex]. The area of the house is [tex]5000[/tex].

[tex](100 - 2w)(100 - w) = 5000[/tex]
[tex]10000 - 100w -200w + 2w^2 = 5000[/tex]
[tex]5000 -300w + 2w^2 = 0[/tex]
[tex]2w^2 - 300w + 5000 = 0[/tex]

Solve by formula....

[tex]\frac{- b \pm \sqrt{b^2 - 4\times a \times c}}{2 \times a}[/tex]

[tex]\frac{300 \pm \sqrt{300^2 - 4\times 2 \times 5000}}{2 \times 2}[/tex]

[tex]\frac{300 \pm \sqrt{90000 - 40000}}{4}[/tex]

[tex]\frac{300 \pm \sqrt{50000}}{4}[/tex]

[tex]\frac{300 \pm 223.61}{4}[/tex]

[tex]\frac{300 + 223.61}{4} = 130.9[/tex]
[tex]\frac{300 - 223.61}{4} = 19.1[/tex]

So width of driveway is [tex]19.1[/tex] feet.

Dimensions of house are............ [tex](100 - 19.1)[/tex] by [tex](100 - 38.2)[/tex]

Dimensions are:- Front to Back = [tex]80.9[/tex] and Width = [tex]61.8[/tex] feet

[Guest]

I was "guessing" how to solve. I didn't know. My apology for wasting so much of your time.

[Guest]

I don't know how you were "guessing" if suddenly you now know all about advanced methods of manipulating algebriac expressions and earlier you were unable to do either any logical reasoning of the problem, or even set out simple expressions using algebra.

[Guest]

I WAS guessing.

[Guest]

I thought it might be similar to the furrows problem. FOIL was used there. I was not sure.

[Guest]

Can you do basic algebra or not.....You need to be able to basic algebra before attempting questions like these.

potato = 0+0+0+0+0+0+0+0 .......do you agree?

If a brick weighs a kilogram plus half a brick, how much does a brick weigh?

[Guest]

"potato = 0+0+0+0+0+0+0+0" I don't understand what you mean.

"If a brick weighs a kilogram plus half a brick, how much does a brick weigh?" 1.5 kilograms.

[Guest]

No, you are not often right and you are wrong again.....

[Guest]

Ok