maths question

[16v_kid]

ahah abit random but i cant figure out how to right an equation using simultaneous equation.

here is the 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?


please help, im really confused about this and i have a test on it on Tuesday.

TIA.

cheers

[GTom]

Perimeter = 40

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

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

so your 2 eqns are:
2x +2y = 40
x = y + 4

put them together and you should get y=8 and x=12

[16v_kid]

ohh thanks alot man.

now i can do it!

i got up to the 2x+2y=40, then i couldt figure out the next equation.

thanks you alot.

cheers

[whoalse]

Wow...I see math tuition here too...? lol...

[16v_kid]

oh and one more equation that's doing my head in.

- One number is 9 less than three times a second number. If the first number plus twice the second number is 16, find the two numbers.



???

[peedman]

x = 3y - 9
x + 2y = 16


x = 16 - 2y
16 - 2y = 3y - 9
5y = 25
y = 5

x + 2(5) = 16
x = 6

x = 6, y = 5

[Treza360]

oh and one more equation that's doing my head in.
- On number is 9 less than three times a second number. If the first number plus twice the second number is 16, find the two numbers.
???

- One number (x) is 9 less than three times a second number (y). If the first number (x) plus twice the second number (y) is 16, find the two numbers.

Similar deal again (assuming your first word is meant to be one):
Eqn. 1: x = 3y - 9
Eqn. 2: x + 2y = 16

Good to see you're studying.

Cheers,
Trent