Equations: 3x-5y = 16 and xy = 7 Solutions: (7, 1) and (-5/3, -21/5)
They are: (3, 1) and (-11/5, -8/5)
Are you sure the equation is not: -3x2 - 2x + 5 = 0? Which is the same as 3x2 + 2x - 5 = 0 which factorises into (3x + 5)(x - 1) with solutions x = 1 and x = -5/3 = -1 2/3 ~= -1.67. As stated there are no real solutions to the equation, only complex solutions: x = -1/3 + i sqrt(14)/3 and x = -1/3 - i sqrt(14)/3
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Indeterminate: possible solutions: x = 1, y = 3; x = 2, y = 1; x = 3, y = -1 etc
Equations: 3x-5y = 16 and xy = 7 Solutions: (7, 1) and (-5/3, -21/5)
They are: (3, 1) and (-11/5, -8/5)
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
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Are you sure the equation is not: -3x2 - 2x + 5 = 0? Which is the same as 3x2 + 2x - 5 = 0 which factorises into (3x + 5)(x - 1) with solutions x = 1 and x = -5/3 = -1 2/3 ~= -1.67. As stated there are no real solutions to the equation, only complex solutions: x = -1/3 + i sqrt(14)/3 and x = -1/3 - i sqrt(14)/3
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It has 2 solutions and they are x = 2 and y = 1 which are applicable to both equations
Indeterminate: possible solutions: x = 1, y = 3; x = 2, y = 1; x = 3, y = -1 etc
it equals 5
This has a number of solutions.... 1) x=2, y=1 2) x=3, y=3 3) x=4, y=5 etc...
4x² - 12x + 9 = 5 → 4x² - 12x + 4 = 0 → x² - 3x + 1 = 0 → x = (-(-3) ±√((-3)² - 4×1×1))/(2×1) → x = (3 ±√5)/2 → x = (3 + √5)/2 or x = (3 - √5)/2
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