Posted by **Ariza** on Saturday, October 24, 2009 at 11:28pm.

Arectangle has perimeter 64cm and area 23cmsquared. Solve the following system of equations to find the rectangle's dimensions.

- Math -
**Ariza**, Saturday, October 24, 2009 at 11:29pm
the equations are:

l= (23/w)

l+w=32

- Math -
**Reiny**, Saturday, October 24, 2009 at 11:42pm
so sub l= (23/w) into l+w = 32

23/w + w = 32

23 + w^2 = 32w

w^2 - 32w + 23 = 0

solve for w, then plug into l=23/w

I got w = .7357

l = 31.2643

check: .7357x31.2643 = 23.001

.7357 + 31.2643 = 32

- Math -
**Anonymous**, Sunday, October 25, 2009 at 12:08am
thankyou !

the second part to the question is.:

solve the system of equations:

x^2 + y^2 = 1

xy = 0.5

- Math -
**Reiny**, Sunday, October 25, 2009 at 12:24am
do it the same way

from xy = .5 = 1/2

y = 1/(2x)

sub into the first

x^2 + 1/(4x^2) = 1

4x^4 + 1 = 4x^2

4x^4 - 4x^2 + 1 = 0

let x^2 = p

then your equation becomes

4p^2 - 4p + 1 = 0

(2p-1)(2p-1) = 0

p = 1/2

so x^2 = 1/2

x = +/- 1/√2

sub back into xy=1/2 to get the y

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