two numbers whose product and sum is equal?

Answer 1

If you're looking only for whole numbers there are
only 2 pairs - 0,0 and 2,2.
But if you open the question up to any rational numbers
then we have the list of pairs as follows:
0,0
2,2
3,3/2
4,4/3
5,5/4
.
.
.
n, n/(n-1) for n=(1,2,...)

Answer 2

if x and y are such numbers, then we must have xy = x + y. Therefore, xy - y = x => y(x -1) = x. If x =1, then y(x-1) = y * 0 =0 <> x =1, so that there's no solution with x =1. If x<>1, we get y = x/(x-1). So there are infinitely many (actually, uncountable many) pairs (x,y) that satisfie your condition. For each x<>1, there's a value of y. So, for example, if x =2, then y = 2/(2-1) = 2 . If you want 2 distinct numbers, set x = 5 and then y = 5/4 = 1.25. If you want integer numbers, then the only solution is (0,0).

We can write y = x/(x-1) = (x-1 +1)/(x-1) = 1 + 1/(x-1). This is the equation of a hyperbole which is assyntotic to the axis x =1.

Answer 3

0 and 2.
0 * 0 = 0
0 + 0 = 0
2 * 2 = 4
2 + 2 = 4

Answer 4

2 & 2
2+2 =4
2*2 = 4

Answer 5

eleven * eleven = 121, eleven+eleven=22 the sole different risk is one million*121 = 121 and that has a sum of 122. If integers are not the sole risk, then the two equations x*z=121 and x+z = y the place y is the sum we would desire to decrease. hence, z = 121/x and y = x + 121/x whilst graphed, this function has a minimum at x=eleven, which propose z = 121/eleven = eleven comparable answer the two way, yet that isn't consistently be the case.

Answer 6

i would say 0 and 0 or 2 and 2

Answer 7

0 and 0.
2 and 2.

If you're asking about two distinct numbers, then
xy=x+y
xy-y=x
(x-1)y=x
y=x/(x-1)

e.g., if x=4, then y=4/3

Answer 8

0,0 and 2,2

Algebrically
xy=x+y
xy-y=x
(x-1)y=x
y=x/(x-1)

where x and y are 2 distinct numbers.
:)

Answer 9

zero and two. zero multiplied by zero , the answer is zero. If you will add it to zero, the answer is zero also. Two multiplied by two, the answer is four. If you will add two to two, the answer is four also.

Answer 10

2+2/2X2