Y/2
2x - y = 2 2x - y = -2 This system is inconsistent -- that is, there is no solution. To help see why, let Z = 2x - y and notice Z can't be 2 and -2 at the same time.
x2 / y = (6)2 /2 = 36/2 = 18
x/y = 5/(y^2) and y/x = 5/(x^2). So x/y + y/x = 5/(y^2) + 5/(x^2), which equals 5x^2/(x^2 y^2) + 5y^2/(x^2 y^2) equals 5(x^2 + y^2)/25 equals (x^2 + y^2)/5. x^2 = 25/y^2, so you get (25/y^2 + y^4/y^2)/5 equals ((25 + y^4)/y^2)/25, which shows that your math teacher is on crack. Seriously, I'm not sure that's true.
-4
y equals 2
Half correct because x = 2 and y = 6
If [ y = x + 2 ], then x is not -1 when y = 5.If [ y = x + 2 ],then when x = -1, y = 1,and when y = 5, x = 3.
2
-1
A pair of simultaneous eq;ns. 3x + y = 4 x + y = 0 Subtract the two eq'ns. This will eliminate 'y' 2x = 4 x = 2 Substitute this value of 'x' into either eq'n for 'y' 2 + y = 0 y = -2 or 3)2) + y = 4 6 + y = 4 y = 4 - 6 y = -2 ( again) . So the answer is ( x,y) = ( 2, -2).
2
2