Two seperate problems
Hello, norwoodjay!
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)
. . 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)