For this problem, introduce a pronumeral and construct a quadratic equation to solve it.  Two numbers differ by 2, but the difference of their reciprocals is `2/15` What are the numbers?

2 Answers

steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

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This is the correct method.  However, when factoring 0 = x^2 + 2x - 15, we get:

0 = (x+5)(x-3)

So, x = 3 and x = -5

Plugging these by into equation 1, we get:

For x = 3, y = 2+3 = 5

So, we have (3,5)

For x = -5, y = 2 + (-5) = -3

So, we have (-5,-3)

These answers would check with the original conditions.

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lemjay | High School Teacher | (Level 3) Senior Educator

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To solve, let x and y be the two numbers, y being the larger number.

Then, express in math form the condition " two numbers differ by 2".


Isolating the y, it becomes:

`y=2+x`   (Let thisbe EQ1)

Next, express in math form the condtion "the difference of their reciprocal is 2/15".

The reciprocal of the two numbers are 1/x and 1/y. Since y is the larger number, then its reciprocal 1/y is smaller compared to 1/x. So the math form of the second condition is:

`1/x -1/y =2/15`

To simplify this, multiply both sides by the LCD of the three fractions which is 15xy.



` `(Let this be EQ2)

Next, substitute EQ1 to EQ2.


`15(2+x) -15x=2x(2+x)`


`30=2x^2 +4x`

Since the resulting equation is quadratic, to solve for values x, set one side equL to zero.


` `To simplify, divide both sides by 2.



Then, factor right side.


Set each factor equalmto zero and solve for x.






Now that the values of are known, plug-in them to EQ1 to get the values of y.

` `When x=3, the value of y is:


When x=-5, the value of y is:


Therefore, there are two sets of numbers that satisfies the given conditions. The numbers are 3 and 5. And the other two numbers are -5 and -3.