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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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