Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
xy = K so K = 21
(x-y) + (xy - 1) = (x - 1)(y + 1)
Xy/y = x
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
Suppose the numbers are x and y Then the sum of their reciprocals is 1/x + 1/y = y/xy + x/xy = (y+x)/xy = 10/20 = 1/2
Looking at the graph of the function can give you a good idea. However, to actually prove that it is even or odd may be more complicated. Using the definition of "even" and "odd", for an even function, you have to prove that f(x) = f(-x) for all values of "x"; and for an odd function, you have to prove that f(x) = -f(-x) for all values of "x".
xy + x = x(y + 1)
The GCF is xy
x+xy=8 xy=-x+8 y=-1+8/x
xy = x ÷x y = 1
Suppose x is an odd number, then x leaves a remainder when divided by 2. That is, x = 2m+1 (for some integer m). Suppose y is an even number, then y is a multiple of 2 so suppose y = 2n (for some integer n). Then x-y = 2m+1 - 2n = 2m-2n + 1 =2(m-n) + 1 Since m and n are integers, then m+n is an integer so that the sum gives 1 more than a multiple of 2. And that is what an odd number is!
Suppose the number, when written is XY. Then sum of digits = 10 implies that X+Y = 10 Also, 20 < XY < 32 implies that X = 2 or 3. Suppose X = 3 Then X+Y = 10 requires Y = 7. But that is not possible because the number cannot be greater than 32. So, X must be 2. And then, X+Y = 10 implies that Y = 8 and so the number is 28.
If x - 3y = 12, we can solve for x in terms of y: x = 12 + 3y. Then, to find xy, we multiply x by y: xy = y(12+3y) = 12y + 3y^2.
It is XY^2 or X*Y*Y