Chapter 7.4, Problem 7P

To determine

To find: The two numbers whose sum is $10$ and sum of their reciprocal is $\frac{5}{12}$ .

Answer to Problem 7P

The two numbers are $6$ and $4$ .

Explanation of Solution

Given information:

The sum of two numbers is $10$ and sum of their reciprocal is $\frac{5}{12}$ .

Formula used:

We will use ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$ .

We will use L.C.M of the denominator.

We will use cross multiplication method.

Calculation:

Consider that the two number is $a$ and $b$ .

Then the sum of two numbers is $\left(a+b\right)=\mathrm{10......}\left(i\right)$ and the sum of their reciprocal is $\frac{1}{a}+\frac{1}{b}=\frac{5}{12}\mathrm{.....}\left(ii\right)$ .

Recall that we will take the L.C.M of the denominator of $\left(ii\right)$ .

Therefore L.C.M of $\left(a,b\right)$ is $ab$ .

Therefore $\frac{b+a}{ab}=\frac{5}{12}$ .

Recall that we will use cross multiplication method.

  $\frac{b+a}{ab}=\frac{5}{12}\Rightarrow 12*\left(b+a\right)=5*ab\Rightarrow 12\left(b+a\right)=5ab\mathrm{.......}\left(iii\right)$ .

Since from $\left(i\right)$ we have $\left(a+b\right)=10$ .

Therefore $\left(iii\right)$ becomes $12*10=5ab\Rightarrow ab=\frac{12*10}{5}=\mathrm{24......}\left(iv\right)$ .

Recall that we will use ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$ .Where $a+b=10$ and $ab=24$ .

Therefore ${\left(a-b\right)}^2={10}^2-\left(4*24\right)\Rightarrow 100-96=4$ .

Therefore ${\left(a-b\right)}^2=4\Rightarrow \left(a-b\right)=2$ .

Now we will solve the expression $\left(a-b\right)=2$ and $\left(a+b\right)=10$ simultaneously.

Therefore we have $a=6$ and $b=4$ whose sum is $10$ and sum of their reciprocal is $\frac{5}{12}$ .