Question

The sum of two numbers is 15 and the sum of their reciprocal is $\dfrac{3}{10}$. Find the numbers. \[\]

Answer 1

Hint: We assume the first number as $x$ and then we get the second number $15-x$. We design the equation using their reciprocals $\dfrac{1}{x}+\dfrac{1}{15-x}=\dfrac{3}{10}$ in accordance with the question. We simplify the equation to obtain a quadratic equation in $x$ which we solve by splitting the middle term method. \[\]

Complete step by step answer:
We know that the multiplicative inverse otherwise known as reciprocal of any real number $a$ is a real number with whom we multiply $a$ to get the multiplicative identity. The reciprocal of $a$ is given by $\dfrac{1}{a}$.\[\]
We know that the quadratic equation in the general form $a{{x}^{2}}+bx+c=0$ ( $a,b,c$ are real numbers and $a\ne 0$ ) can be solved by splitting the middle term into two numbers $p,q$ such that $p,q$ will be factors of $c\times a$ and their addition or subtraction will be $b$. \[\]
Let the first number be $x$ , then the second number be $y$. So the reciprocals of the first number is $\dfrac{1}{x}$ and the reciprocal of $y$ is $\dfrac{1}{y}$. We are given in the question that the sum of two numbers is 15 We have
\[\begin{align}
  & x+y=15 \\
 & \Rightarrow y=15-x \\
\end{align}\]
So the two numbers can be expressed in terms of $x$ as $x,15-x$ whose reciprocals are $\dfrac{1}{x},\dfrac{1}{15-x}$. We are further given the question that the sum of their reciprocal is $\dfrac{3}{10}$. So we have,
\[\dfrac{1}{x}+\dfrac{1}{15-x}=\dfrac{3}{10}\]
We multiply $x\left( 15-x \right)$ in all the terms appearing in above equation to have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}\cdot x\left( 15-x \right)+\dfrac{1}{15-x}x\left( 15-x \right)=\dfrac{3}{10}x\left( 15-x \right) \\
 & \Rightarrow 15-x+x=\dfrac{3\left( 15x-{{x}^{2}} \right)}{10} \\
 & \Rightarrow 15=\dfrac{3\left( 15x-{{x}^{2}} \right)}{10} \\
\end{align}\]
We cross-multiply to have,
\[\begin{align}
  & \Rightarrow 15x-{{x}^{2}}=15\times \dfrac{10}{3}=50 \\
 & \Rightarrow {{x}^{2}}-15x+50=0 \\
\end{align}\]
We compare the above quadratic equation in $x$ to general equation $a{{x}^{2}}+bx+c=0$ to have $a=1,b=-15,c=150$. We split the middle term $-15x$ into $-10x,-5x$ as $-10+\left( -5 \right)=-15=b$ and $-5\times \left( -10 \right)=50=c\times a=50\times 1$. So we have
\[\begin{align}
  & \Rightarrow {{x}^{2}}-10x-5x+50=0 \\
 & \Rightarrow x\left( x-10 \right)-5\left( x-10 \right)=0 \\
 & \Rightarrow \left( x-10 \right)\left( x-5 \right)=0 \\
 & \Rightarrow x-10=0\text{ or }x-5=0 \\
 & \Rightarrow x=10\text{ or }x=5 \\
\end{align}\]

So we have obtained two roots of the equation $x=10,5$. If we take $x=10$ as the first number then the second number is $15-x=15-10=5$. If we take $x=5$ as the first number then the second number is $15-x=15-5=10$. So the two numbers are 5 and 10.\[\]

Note: We note that the reciprocal of 0 does not exist. We can add the fractional expressions $\dfrac{1}{x},\dfrac{1}{15-x}$ using the least common multiple of denominators. We can find the splitting terms for the quadratic equation using the prime factorization of $c\times a$. We can alternatively find the roots of the quadratic equation by completing square method and the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.