One positive integer is 3 units more than another.

Question

Could someone please work this out for me? I have a really tough time with word problems and this one has me stuck.

 

 

One positive integer is 3 units more than another. When the reciprocal of the larger is subtracted from twice the reciprocal of the smaller, the result is 2/9. Find the two positive integers.

Answer 1

Marisa,

 

Let you 1st integer = x and the 2nd integer = y where y = x + 3.

So y > x.

 

The 2nd sentence tells us that:

 

2(1/x) - 1/y = 2/9

 

Substitute y = x+3

 

2/x - 1/(x+3) = 2/9

 

The least common denominator on the left side is x(x+3) so we can rewrite that side as:

 

[2(x+3) - (x)]/[x(x+3)] = 2/9

 

Multiply both sides by x(x+3) to get:

 

2(x+3) - x = 2x(x+3)/9

⇒

2x+6-x = x+6 = 2x(x+3)/9

 

Now multiply both sides by 9:

 

9x+54 = 2x²+6x

 

Combining terms gives:

 

2x²-3x-54 = 0

 

We can factor this as:  (2x+9)(x-6) = 0

 

Which gives us 2 solutions:  x = -9/2, and x = 6

 

The question asks us to pick 2 positive integers so:

 

x = 6 & y =9

 

Check:  2(1/6) - 1/9 = 2/6 - 1/9 = 1/3 - 1/9 = 3/9 - 1/9 = 2/9