find x3+y2 if x=4 and y=3 ( 4*4*4)+(3*3) 64 +9 =73
If x=4... then 'y' MUST equal 3 !
x = 3 y = 4 2x + y - 1 = (6) + (4) - 1 = 9
x = -1/3 and y = 7/12
(3, 2)
x=4 y=x+3 y=4+3 y=7
x+y=-8 2x-y=4 y=-x-8 2x+x+8=4 3x=-4 x=-4/3 -4/3 + y= -8 y = -8 + 4/3 y= -6 2/3
find x3+y2 if x=4 and y=3 ( 4*4*4)+(3*3) 64 +9 =73
equation*6x-3*...answer*6(-4)-3=-27*...variable*x=-4*
If x=4... then 'y' MUST equal 3 !
x=7 y=4
y/x is a constant; so first of you can solve for this constant. y/x=6/-3, or y/x=-2 y/4=-2 y=-8
The value is 1 if (x, y) = (4, 3) and 0 otherwise.
x equals -12 and y equals 1/4 of -12, so y = -3.
x = 3 y = 4 2x + y - 1 = (6) + (4) - 1 = 9
y = x + 2 y = -x + 4 x + 2 = -x + 4 2x + 2 = 4 2x = 2 x = 1 y = x + 2 y = 1 + 2 y = 3 (1, 3)
X = 2 and Y = 3.