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?
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.
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.
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.