A) A rectangle has perimeter 64 cm and area 23 cm^2. Solve the following system of equations to find the rectangle's dimensions.
length=23/width
length + width = 32
Let
L = length
W = width
L=23/W
L+W=32
23/W+W=32
W2-32W+23=0
Solve for the quadraticequation in W.
Find L using L = 23/W.
You should get 0.7 and 31 approx.
B. Solve
x^2+y^2=1
xy=0.5
If you plot the two equations on sqaure paper, you will see a circle (firt equation) and two disjoint curves with asymptotes toward the 4 extremities of the axes. This will give you an idea where the roots will be, namely where they intersect, and that there should be four of them if they cut the circle, zero if they don't, and two (visually) if they are tangential to the circle.
Substitute y=0.5/x into the equation of the circle (with unit radius).
x2 + (0.5/x)2 = 1
substitute u=x2,
u+0.25/u=1
u2 -u + 0.25 = 0
Solving for u, we get
u=0.5 ± 0 (meaning the intersection is tangential)
=0.5
x=±sqrt(0.5)
y=0.5/x=±sqrt(0.5)
A) A rectangle has perimeter 64 cm and area 23 cm^2. Solve the following system of equations to find the rectangle's dimensions.
length=23/width
length + width = 32
B) Solve the system of equations
x^2+y^2=1
xy=0.5
1 answer