# A positive number n is a fraction less than one. Twice n plus the reciprocal of n equals 3. What...

## Question:

A positive number n is a fraction less than one. Twice n plus the reciprocal of n equals 3. What is the value of n?

## Verbal Statement Into the Equation:

We can translate a verbal statement into an equation by using algebraic operations. Here, we use + for "more than", - for "less than, multiplication for "times" etc. We can then solve it using the algebraic operations on both sides of the equation.

The problem says, "twice {eq}n {/eq} plus the reciprocal of {eq}n {/eq} equals {eq}3 {/eq}".

So we get the equation:

$$2n+ \dfrac{1}{n}=3\\[0.4cm] \text{Multiply each term on both sides by n},\\[0.4cm] 2n^2+1= 3n\\[0.4cm] \text{Subtracting 3n from both sides},\\[0.4cm] 2n^2-3n+1=0$$

Here, {eq}\text{(First number) (Second number)}=(2)(1)=2 {/eq} and the middle number = {eq}-3 {/eq}.

We have to find two numbers whose product is {eq}2 {/eq} and whose sum is {eq}-3 {/eq}.

Two such numbers are {eq}-2 {/eq} and {eq}-1 {/eq}.

So we get:

$$2n^2-2n-1n+1=0 \\[0.4cm] (2n^2-2n)+(-1n+1)=0 \\[0.4cm] 2n(n-1) -1 (n-1)=0\\[0.4cm] (n-1)(2n-1)=0\\[0.4cm] n-1=0;\,\,\,2n-1=0\\[0.4cm] n=1; \,\,\, n= \dfrac{1}{2}$$

But the problem says, {eq}n {/eq} is a fraction less than one.

So n cannot be {eq}1 {/eq}.

Therefore, {eq}\color{blue}{\boxed{\mathbf{n= \dfrac{1}{2}}}} {/eq}.