**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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