Let's call the two numbers [tex]s[/tex] and [tex]l[/tex].
Given these variables, we can say: [tex]s = l - 14[/tex], based on the first sentence in the problem.
Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:
[tex]\dfrac{1}{s} + 5\Big( \dfrac{1}{l} \Big) = \dfrac{1}{4}[/tex]
To solve, we can simply substitute [tex]l -14[/tex] in for [tex]s[/tex] in the second equation and solve.
[tex]\dfrac{1}{l - 14} + \dfrac{5}{l} = \dfrac{1}{4}[/tex]
[tex]\dfrac{l}{l(l - 14)} + \dfrac{5(l - 14)}{l(l - 14)} = \dfrac{1}{4}[/tex]
[tex]\dfrac{l + 5l - 70}{l(l - 14)} = \dfrac{1}{4}[/tex]
[tex]4(6l - 70) = l(l - 14)[/tex]
[tex]24l - 280 = l^2 - 14l[/tex]
[tex]l^2 - 38l + 280 = 0[/tex]
[tex](l - 10)(l - 28) = 0[/tex]
[tex]l = 10, 28[/tex]
Now, notice that we have found two solutions, but the problem is only asking for one. This likely means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,
[tex]s = 10 - 14 = -4[/tex],
Since [tex]s[/tex] is not positive in this case, [tex]l = 10[/tex] is not a solution.
Thus, [tex]l = 28[/tex] is our only solution. In this case,
[tex]s = 28 - 14 = 14[/tex],
which means that the smaller number is 14 and the larger number is 28.
