If Y = 0 then there is no value of X such that XY = 1.
maybe
So far you know that: x=-8 xy=1 so xy means x times y. If you split xy. y=1/x or y=1/-8 1/-8= -0.125 You can find out that this is correct with the formula xy=1 because -0.125 x -8 is 1.
x+xy=8 xy=-x+8 y=-1+8/x
Suppose x + y + z = 0 then z = - x - y = -(x + y) . . . . . . (A) 1/x + 1/y + 1/z = 0 implies x, y and z are all non-zero: otherwise the reciprocals are undefined. then z ≠0 implies that x+y ≠0 (by (A)) so 1/x + 1/y + 1/[-(x+y)] = 0 (using (A)) so that 1/x + 1/y = 1/(x+y) ie (x + y)/xy = 1/(x + y) Now, since x+y ≠0, multiply both sides by x+y to give (x + y)2/xy = 1 or (x + y)2 = xy x2 + 2xy + y2 = xy x2 + xy + y2 = 0 so that x = [-y ± sqrt(y2 - 4y2)]/2 = [-y ± y*sqrt(-3)]/2 which is cannot be real if y is real.
If Y = 0 then there is no value of X such that XY = 1.
"xy" refers to the product of "x" and "y". Just substitute the variables for numbers, and do the multiplication.
xy = x ÷x y = 1
maybe
So far you know that: x=-8 xy=1 so xy means x times y. If you split xy. y=1/x or y=1/-8 1/-8= -0.125 You can find out that this is correct with the formula xy=1 because -0.125 x -8 is 1.
x+xy=8 xy=-x+8 y=-1+8/x
Suppose x + y + z = 0 then z = - x - y = -(x + y) . . . . . . (A) 1/x + 1/y + 1/z = 0 implies x, y and z are all non-zero: otherwise the reciprocals are undefined. then z ≠0 implies that x+y ≠0 (by (A)) so 1/x + 1/y + 1/[-(x+y)] = 0 (using (A)) so that 1/x + 1/y = 1/(x+y) ie (x + y)/xy = 1/(x + y) Now, since x+y ≠0, multiply both sides by x+y to give (x + y)2/xy = 1 or (x + y)2 = xy x2 + 2xy + y2 = xy x2 + xy + y2 = 0 so that x = [-y ± sqrt(y2 - 4y2)]/2 = [-y ± y*sqrt(-3)]/2 which is cannot be real if y is real.
It equals y+1
y=f(x)= x(2x+y)=7 f(x)=2x^2 +xy -7 = 0 y=2x^2 +xy -7 y-xy -7 + 2x^2 = 0 y(1-x)=2x^2-7 y=(2x^2-7) / (1-x) Excuse the working. Y equals 2 X squared minus 7 all divided by 1-X
0
-1/8 -8*-1/8 = 1
There are no intercepts because the curve, xy = 4 is asymptotic. When x = 0 (where the y intercept would be) y is infinite, and conversely, when y = 0 x is infinite.