2x-y+4=0... what are you actually trying to find?
2xy - 4x plus 8y - 16 equals x plus 4 in parentheses multiplied by 2y minus 4 in parentheses. So, the factors are ( x + 4) and ( 2y - 4) .
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
22 = 4
x = y = 0 ?
That would still be 4. If it were times 0 it would be 0.
2xy - 4x plus 8y - 16 equals x plus 4 in parentheses multiplied by 2y minus 4 in parentheses. So, the factors are ( x + 4) and ( 2y - 4) .
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
22 = 4
Assuming y' is dy/dx, y = x^4/4 + yx^2
x = y = 0 ?
x + 3y = 4 -x + 2y = -4 Simultan eous eq'ns. Eliminate 'x' by adding. 4 Hence 5y = 0 y = 0 When y = 0 Substitute x + 3(0) = 4 x + 0 = 4 x = 4 So the answer as a coordinate pairs is ( x,y) = ( 4,0)
the difference between 2x2 +4xy-3 and x2-2xy-4 is?
(2xy-4x)+ (8y-16)=2xy-4x+8y-16, which factors as 2(x+4)(y-2)
4(x + 1) = 0.
That would still be 4. If it were times 0 it would be 0.
x = 0
4xy - 2y(x + 4) = 4xy - 2xy - 8y = 2xy - 8y = 2y(x - 4)