# 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?

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### 2 Answers

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.

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".

`y-x=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.

`15xy(1/x-1/y)=2/15*15xy`

`15y-15x=2xy`

` `(Let this be EQ2)

Next, substitute EQ1 to EQ2.

`15y-15x=2xy`

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

`30+15x-15x=4x+2x^2`

`30=2x^2 +4x`

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

`0=2x^2+4x-30`

` `To simplify, divide both sides by 2.

`0/2=(2x^2+4x-30)/2`

`0=x^2+2x-15`

Then, factor right side.

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

Set each factor equalmto zero and solve for x.

`(x-3)=0`

`x=3`

and

`x+5=0`

`x=-5`

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:

`y=2+x=2+3)=5`

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

`y=2+x=2+(-5)=2-5=-3`

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.