# Find the value of x and y. - Simultaneous Equationhttp://goo.gl/pBCu6 Given that the region planted with orchids is 460 m^2 and the perimeter of the rectangular fish pond is 48 m, find the value of...

Find the value of x and y. - Simultaneous Equation

http://goo.gl/pBCu6

Given that the region planted with orchids is 460 m^2 and the perimeter of the rectangular fish pond is 48 m, find the value of x and y.

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The picture suggests the dimensions of the fish pond: Length = `30 - y` and the width = `x - 10` .

The perimeter of fish pound is of 48 m such that:

`2(L+w) = P =gt 2(30 - y + x - 10) = 48 =gt 20 + x - y = 24 =gt x - y = 24 - 20 =gt x - y = 4 =gt x = 4+y`

Looking at the picture you should see two rectangular regions planted with orchids whose dimensions are:

region 1 - length=x; width=y

region 2- length=10; width=30-y

Evaluating the total area of region planted with orchids yields:

A = Area of region 1 + Area of region 2

`A =xy + 10*(30-y) =gt 460 = 300- 10y+ xy`

`xy - 10y= 460 - 300`

`xy - 10y = 160`

Plugging `x = y+4` in equation `xy - 10y = 160` yields:

`y(y + 4) -10y = 160`

Opening the brackets yields:

`y^2 - 6y - 160= 0`

`y_(1,2) = (6+-sqrt(36+ 640))/2 =gt y_1 = (6+26)/2 =gt y_1 = 16 m`

`y_2 = (6-26)/2 = -10`

You need to exclude the value `y_2 = -10` because it is not possible for a length to be negative.

x = 4 + 16 => x = 20 m

**Hence, evaluating the values of x and y yields: x = 20 m ; y = 16 m.**