Tuesday, March 19, 2019

Algebra: Solving Simple Non-Linear Systems

Algebra: Solving Simple Non-Linear Systems




System I:  

x + y = a
x^2 + y = b

Solving for y:
x + y = a
y = a - x

Subtracting the two equations from the system:
x + y = a
- [x^2 + y] = -[ b ]

x - x^2 = a - b
x^2 - x = b - a
x^2 - x - (b - a) = 0

Solving for x:
x = ( 1 ± √(1 - 4*(b - a) ) / 2

Summary for System I:
x = ( 1 ± √(1 - 4*(b - a) ) / 2
y = a - x

If a and b are real numbers, then 1 - 4*(b - a) ≥ 0, and
1 ≥ 4*(b - a)

System II:

x + y = a
x + y^2 = b

Solving for x:
x + y = a
x  = a - y

Subtracting the two equations from the system:
x + y = a
- [ x + y^2 ] = -[ b ]

y - y^2 = a - b
y^2 - y = b - a
y^2 - y - (b - a) = 0

Solving for y:
y = ( 1 ± √(1 - 4*(b - a) )/2

Summary for System II:
x  = a - y
y = ( 1 ± √(1 - 4*(b - a) )/2

System III:

x + y = a
x^2 + y^2 = b

Solving for y:
y = a - x

Solving for x:
x^2 + (a - x)^2 = b
x^2 + a^2 - 2*a*x + x^2 = b
2*x^2 - 2*a*x + (a^2 - b) = 0

x = ( 2*a ± √(4*a^2 - 4*2*(a^2 - b) ) / 4
x = ( 2*a ± √(4*a^2 - 8*(a^2 - b) ) / 4
x = ( 2*a ± √(4*a^2 - 8*a^2 + 8*b) ) / 4
x = ( 2*a ± √(8*b - 4*a^2) ) / 4
x = ( a ± √(2*b - a^2) ) / 2

Summary for System III:
x = ( a ± √(2*b - a^2) ) / 2
y = a - x

System IV:

x^2 + y^2 = a
x * y = b

Solving for y:
y = b / x 

I'm assuming that x ≠0 and y ≠0.

x^2 + y^2 = a
x^2 + (b / x)^2 = a
x^4 + b^2 = a * x^2
x^2 - a * x^2 + b^2 = 0

Let w = x^2, then w^2 = x^4

Then:
w^2 - a*w + b^2 = 0

Then:
w = (a ± √(a^2 - 4 * b^2) )/ 2

And:
x = ± √( (a ± √(a^2 - 4 * b^2) )/ 2 )

We have four answers to the system.

Summary for System IV:
x = ± √( (a ± √(a^2 - 4 * b^2) )/ 2 )
y = b / x 

A lot of fun,

Eddie

All original content copyright, © 2011-2019.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

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