The sume of the reciprocals of two consecutive integers is 3/2. Find the two integers

1 and 2

Step-by-step explanation:

Let's say our first number is x. Our second number is the number after that, or x+1. A reciprocal of a number is 1/that number. Therefore, the reciprocal of x is 1/x and the reciprocal of x+1 is 1/(x+1). The question states the sum of the reciprocals of the two numbers is equal to 3/2. Therefore,

1/x + 1/(x+1) = 3/2

multiply both sides by 2 to remove a denominator

2/x + 2/(x+1) = 3

multiply both sides by x to remove a denominator

2 + (2*x)/(x+1) = 3x

multiply both sides by (x+1) to remove the other denominator

2*(x+1) + 2*x = 3x*(x+1)

expand

2*x+2 + 2*x = 3x²+3

combine like terms

4x + 2 = 3x²+3

subtract (4x+2) from both sides to make everything equal to 0 and to form a quadratic

3x² - 4x + 1 = 0

To factor this, we need to find two numbers that add up to b and multiply to a*c in an equation of form ax²+bx + c. Here, a=3, c=1, and b = -4

Two numbers that add to -4 and multiply to 3*1=3 are -3 and -1. We can thus factor this out to get

3x²-3x-x+1 = 0

3x(x-1) -1 ( x-1) = 0

(3x-1)(x-1) = 0

Therefore, solving for 0, we get

3x-1=0

x = 1/3

x-1 = 0

x=1

The only integer solution possible is x=1

To confirm, 1/1 + 1/(1+1) = 3/2, so x=1 is correct, with 1+1=2 being the second integer

is this a multiple choice question?

if so it's a and c

answer: sorry i can't you.

step-by-step explanation: