# YR10 Simaltaneous Equations

• Oct 10th 2009, 09:55 PM
ktmboy
YR10 Simaltaneous Equations
Hello,

I cant figure out how to how to write an equation for this question.

-A rectangular house has a perimeter of 40 meters and the length is 4 meters more than the width. What are the dimensions of the house?

TIA.

cheers
• Oct 10th 2009, 10:11 PM
Glavata
Let l = the length, w = the width.
If the length is 4 more than the width, then l = w+4 | w we do not know.
The parameter is given to be 40 so, you have 2 lengths and 2 widths.
w + (w+4) + w + (w+4) = 40
4w+8 = 40
4w = 32
w = 8

So the width (w) is 8, since the length (l) is 4 more the width (w) it must be 12. 8 + 12 + 8 12 = 40
• Oct 10th 2009, 10:15 PM
ktmboy
i was looking for something like this

Perimeter = 40

x (length), y (width)
therefore eqn(1) = 2x +2y = 40

Length is 4 more than width
therefore eqn(2) = x = y + 4

2x +2y = 40
x = y + 4

put them together and you should get y=8 and x=12
• Oct 10th 2009, 10:22 PM
mr fantastic
Quote:

Originally Posted by ktmboy
i was looking for something like this

Perimeter = 40

x (length), y (width)
therefore eqn(1) = 2x +2y = 40

Length is 4 more than width
therefore eqn(2) = x = y + 4