the equations are:
l= (23/w)
l+w=32
Arectangle has perimeter 64cm and area 23cmsquared. Solve the following system of equations to find the rectangle's dimensions.
4 answers
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
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
thankyou !
the second part to the question is.:
solve the system of equations:
x^2 + y^2 = 1
xy = 0.5
the second part to the question is.:
solve the system of equations:
x^2 + y^2 = 1
xy = 0.5
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
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