• Nov 15th 2006, 05:46 AM
Fnus
Hi, this might be really simple, but I just don't get it, so I'd like some help ^^

Question 1:
'Part of a garden consist of a square lawn with a path 1.5 metres wide around its perimeter. If the lawn area is two-thirds of the total area, find the lenght of a side of the lawn.'

Question 2:
Solve the simultanious equations:
xy= 42
x - 4y = 17

Question 3:
Solve the equation:
(I don't know how to write fractions, but it's fractions..)

1/x + 2 + 1/ x^2 - 4 = 2/5

Help is very appreciated. ^^
• Nov 15th 2006, 12:25 PM
TriKri
Quote:

Originally Posted by Fnus
Hi, this might be really simple, but I just don't get it, so I'd like some help ^^

Question 1:
'Part of a garden consist of a square lawn with a path 1.5 metres wide around its perimeter. If the lawn area is two-thirds of the total area, find the lenght of a side of the lawn.'

Question 2:
Solve the simultanious equations:
xy= 42
x - 4y = 17

Question 3:
Solve the equation:
(I don't know how to write fractions, but it's fractions..)

1/x + 2 + 1/ x^2 - 4 = 2/5

Help is very appreciated. ^^

The fisrt problem is answerd by setting up solving the equation $\displaystyle x^2 = \frac{2}{3} * (x + 1.5 m)^2$. It's a second order equation.

To solv the second problem, you start this way:

Allway isolate one of the unknown variables so you get an equation for it. In this case, you can isolate y by dividing the first equation by x. You'll get
$\displaystyle y= 42/x$
which also tells you that x isn't zero.
By insertion into the second equation you'll get
$\displaystyle x - 4*42/x = 17$
$\displaystyle x^2 - 84 = 17x$
That's a second order equation too. But I guess you can see that stuff for yourself ;)
• Nov 15th 2006, 12:28 PM
TriKri
The fisrt problem is answerd by setting up solving the equation $\displaystyle x^2 = \frac{2}{3} * (x + 1.5 m)^2$. It's a second order equation.

To solv the second problem, you start this way:

Allway isolate one of the unknown variables so you get an equation for it. In this case, you can isolate y by dividing the first equation by x. You'll get
$\displaystyle y= 42/x$
which also tells you that x isn't zero.
By insertion into the second equation you'll get
$\displaystyle x - 4*42/x = 17$
$\displaystyle x^2 - 84 = 17x$
That's a second order equation too. But I guess you can see that stuff for yourself ;)

The third, multiply the equation by x^2 and you'll get a second order equation here too.