Give all solutions of each nonlinear systems of equations, including those with nonreal complex components.

x2+y2=10 x2+y2=4

2x2 -y2=17 2x-3y2=-12

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- July 23rd 2008, 04:11 PMnorwoodjayI need help / complete solutions for 2 problems
Give all solutions of each nonlinear systems of equations, including those with nonreal complex components.

x2+y2=10 x2+y2=4

2x2 -y2=17 2x-3y2=-12 - July 23rd 2008, 06:06 PMnorwoodjay
Two seperate problems

- July 23rd 2008, 06:12 PMtutor
Hi,

solving the equation here

adding the equations here u will get

3x^2=27

x^2=9

x=3

and substitute x=3

y^2=10-9

y=1

for second answer

multiplying 3 in first part then add both the equation

5x^2=0

x=0

when x=0 in first part then y=2

(Nerd) - July 23rd 2008, 07:54 PMSoroban
Hello, norwoodjay!

Quote:

Solve the systems of equations:

. . .x² + y² .= .10 . [1]

. . 2x² - y² .= .17 . [2]

Add [1] and [2]: .3x² = 27 . → . x² = 9 . → . x = ±3

Substitute into [1]: .(±3)² + y² .= .10 . → . y² = 1 . → . y = ±1

Solutions: .(3,1), (3,-1), (-3,1), (-3,-1)

Quote:

. . x² + y² .= . 4 .[1]

. .2x - 3y² .= -12 .[2]

Multiply [1] by 3: .3x² + 3y² .= .12

. . . . . . Add [2]: . 2x .- .3y² .= -12

And we have: .3x² + 2x .= .0 . → . x(3x + 2) .= .0

. . Hence: .x .= .0, -2/3

Substitute into [1] and we get: .y .= .±2, ±4√2/3

Solutions: . (0, ±2), .(-2/3, ±4√2/3)

- July 24th 2008, 02:02 AMnorwoodjay
Thanks alot.