7+xy
x*(y+z) = x*y + x*z This is the distributive property of multiplication over addition.
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
x = 10, y = 25
6x - (6 + y)
2(x+y) is twice the sum of x and y, and 2x+y is the sum of twice x and y
The sum of x and y decreased by their product is (x + y)- xy.
3x over x + y
1/x + 1/y = (y+x)/xy But y + x = sum = 150, and xy = product = 40 So sum of reciprocals = 150/40 = 3.75
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. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
x*(y+z) = x*y + x*z This is the distributive property of multiplication over addition.
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
x = 10, y = 25
Twice the sum of 'x' and 'y' . . . 2(x+y) The sum of twice 'x' and 'y' . . . (2x+y)
6x - (6 + y)
2(x+y) is twice the sum of x and y, and 2x+y is the sum of twice x and y
x + y = 28 x*y = 7 x=7/y replacing X in eq 1 7/y+y=28 y^2 - 28y + 7 = 0 using above solutions find the reciprocal and sum -- Dhruv