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

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