Question

Sum of two natural numbers is 8 and the difference of their reciprocals is $\bold{\frac{2}{15} }$ . Find the numbers?? Please thank my answers ❤️❤️❤️​

Answer 1

Answer:-

$\red{\bigstar}$ The numbers are $\large\leadsto\boxed{\rm\purple{3 \: and \: 5}}$

• Given:-

• Sum of two natural numbers is 8.

• Difference of their reciprocals is 2/15

• To Find:-

• The numbers

• Solution:-

• Let the numbers be 'x' and 'y'.

• According to the question:-

✴ $\sf x + y = 8 \dashrightarrow\bf\red{[eqn.i]}$

✴ $\sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \dashrightarrow\bf\red{[eqn.ii]}$

• Taking eqn[i]:-

$\longrightarrow \sf x + y = 8$

$\longrightarrow \sf x = 8 - y$

• Substituting this value of x in eqn[ii]:-

$\\ \longrightarrow \sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{1}{(8-y)} - \dfrac{1}{y} = \dfrac{2}{15}$

• Taking LCM:-

$\\ \longrightarrow \sf \dfrac{y - (8-y)}{y(8-y)} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{y - 8 + y}{8y-y^2} = \dfrac{2}{15}$

$\\ \longrightarrow \sf \dfrac{2y - 8}{8y-y^2} = \dfrac{2}{15}$

• Cross multiplying:-

$\\ \longrightarrow \sf 15 \times(2y-8) = 2 \times (8y-y^2)$

$\\ \longrightarrow \sf 30y - 120 = 16y - 2y^2$

$\\ \longrightarrow \sf 30y - 16y - 120 = - 2y^2$

$\\ \longrightarrow \sf 2y^2 - 14y - 120$

$\\ \longrightarrow \sf y^2 - 7y - 60$

• Splitting the middle term:-

$\\ \longrightarrow \sf y^2 - 5y + 12y - 60$

$\\ \longrightarrow \sf y(y - 5) + 12(y - 5)$

$\\ \longrightarrow \sf (y-5)(y+12)$

Now,

$\\ \bigstar \: \sf y - 5 = 0$

$\\ \longrightarrow \sf y = 5$

$\\ \bigstar \: \sf y + 12 = 0$

$\\ \longrightarrow \sf y = -12$

Here, y = -12 is not possible.

Hence,

$\pink{\bigstar}$ $\bf\green{y = 5}$

• Substituting the value of y in eqn[i]:-

$\\ \longrightarrow \sf x + y = 8$

$\\ \longrightarrow \sf x + 5 = 8$

$\\ \longrightarrow \sf x = 8 - 5$

$\pink{\bigstar}$ $\bf\green{x = 3}$

Therefore, the numbers are 3 and 5.

Answer 2

Answer :-

$\red{\bigstar}$ The numbers are $\large\leadsto\boxed{\rm\purple{3 \: and \: 5}}$

• Given :-

• Sum of two natural numbers is 8.
• Difference of their reciprocals is 2/15.

• To Find :-

• The numbers.

• Solution :-

• Let the numbers be 'x' and 'y'.

• According to the question:-

✴ $\sf x + y = 8 \dashrightarrow\bf\red{[eqn.i]}$

✴ $\sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \dashrightarrow\bf\red{[eqn.ii]}$

• Taking eqn[i]:-

$\longrightarrow \sf x + y = 8$

$\longrightarrow \sf x = 8 - y$

• Substituting this value of x in eqn[ii]:-

$\begin{gathered}\\ \longrightarrow \sf \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \\ \end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf \dfrac{1}{(8-y)} - \dfrac{1}{y} = \dfrac{2}{15}\end{gathered}$

• Taking LCM:-

$\begin{gathered}\\ \longrightarrow \sf \dfrac{y - (8-y)}{y(8-y)} = \dfrac{2}{15} \\ \end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf \dfrac{y - 8 + y}{8y-y^2} = \dfrac{2}{15}\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf \dfrac{2y - 8}{8y-y^2} = \dfrac{2}{15}\end{gathered}$

• Cross multiplying :-

$\begin{gathered}\\ \longrightarrow \sf 15 \times(2y-8) = 2 \times (8y-y^2)\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf 30y - 120 = 16y - 2y^2\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf 30y - 16y - 120 = - 2y^2\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf 2y^2 - 14y - 120\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf y^2 - 7y - 60\end{gathered}$

• Splitting the middle term:-

$\begin{gathered}\\ \longrightarrow \sf y^2 - 5y + 12y - 60\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf y(y - 5) + 12(y - 5)\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf (y-5)(y+12)\end{gathered}$

Now,

$\begin{gathered}\\ \bigstar \: \sf y - 5 = 0\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf y = 5\end{gathered}$

$\begin{gathered}\\ \bigstar \: \sf y + 12 = 0\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf y = -12\end{gathered}$

Here, y = -12 is not possible.

Hence,

$\pink{\bigstar}$ $\bf\green{y = 5}$

• Substituting the value of y in eqn[i]:-

$\begin{gathered}\\ \longrightarrow \sf x + y = 8\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf x + 5 = 8\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf x = 8 - 5\end{gathered}$

$\pink{\bigstar}$ $\bf\green{x = 3}$

Therefore, the numbers are 3 and 5.