Question

Solve $\dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b}$ for $f$ and then find the value of $f?$

Answer 1

Hint: In this question, we are going to solve for $f$ and find the value of $f$.
First, we are going to make the right hand side of the equation a single fraction by finding a common denominator to the fraction.
Substitute the obtained fraction in the given equation and ten multiply and divide variables wherever required to get the solution.
Hence, we can get the required result.

Complete step-by-step solution:
In this question, we are going to solve and find the value of $f$.
First write the given equation and mark it as $\left( 1 \right)$
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b}...\left( 1 \right)$
When we solve for $f,$ we should isolate $f$ on one side of the equation.
Now we need to make the right hand side of the equation a single fraction.
We do this by finding a common denominator.
Take LCM on the right hand side of the fraction to make the denominator the same.
$ \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{b}{{ab}} + \dfrac{a}{{ab}}$
$ \Rightarrow \dfrac{{b + a}}{{ab}}$
$ \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{{a + b}}{{ab}}...\left( 2 \right)$
Now substitute equation $\left( 2 \right)$in equation $\left( 1 \right)$ we get,
We have $\dfrac{1}{f} = \dfrac{{a + b}}{{ab}}$
Multiply both sides by $f$ to the above fraction we get,
$ \Rightarrow \left( f \right)\dfrac{1}{f} = \left( f \right)\dfrac{{a + b}}{{ab}}$
$f$ Get cancelled on the left hand side of the equation.
$ \Rightarrow 1 = \left( f \right)\dfrac{{a + b}}{{ab}}$
Now multiply both sides by$ab$, we get
$ \Rightarrow ab = \left( f \right)\dfrac{{a + b}}{{ab}}\left( {ab} \right)$
$ab$ Get cancelled on the right hand side of the equation.
$ \Rightarrow ab = f\left( {a + b} \right)$
Finally, divide both sides by$a + b$, we get
$ \Rightarrow \dfrac{{ab}}{{a + b}} = \dfrac{{f\left( {a + b} \right)}}{{a + b}}$
$ \Rightarrow \dfrac{{ab}}{{a + b}} = f$

Thus the value of $f$ is equal to $\dfrac{{ab}}{{a + b}}$

Note: We can also solve this fraction by the following method:
First, subtract $\dfrac{1}{b}$from each side of the equation to isolate the $f$ term by keeping the equation balanced.
$ \Rightarrow \dfrac{1}{f} - \dfrac{1}{b} = \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{b}$
$ \Rightarrow \dfrac{1}{f} - \dfrac{1}{b} = \dfrac{1}{a} + 0$
$ \Rightarrow \dfrac{1}{f} - \dfrac{1}{b} = \dfrac{1}{a}$
Next, subtract the fractions on the right side of the equation after putting each fraction over a common denominator by multiplying each fraction by appropriate form.
$ \Rightarrow \dfrac{1}{a} = \left( {\dfrac{b}{b} \times \dfrac{1}{f}} \right) - \left( {\dfrac{1}{b} \times \dfrac{f}{f}} \right)$
Let us rewriting we get
$ \Rightarrow \dfrac{1}{a} = \left( {\dfrac{b}{{bf}}} \right) - \left( {\dfrac{f}{{bf}}} \right)$
Then we get,
$ \Rightarrow \dfrac{1}{a} = \dfrac{{b - f}}{{bf}}$
Flip the fraction on each side of the equation to keep the equation balanced.
$ \Rightarrow \dfrac{a}{1} = \dfrac{{bf}}{{b - f}}$
On writing we get
$ \Rightarrow a = \dfrac{{bf}}{{b - f}}$
Multiply each side of the equation by $abf$
$ \Rightarrow abf \times \dfrac{1}{a} = abf \times \dfrac{{b - f}}{{bf}}$
On cancel the term and we get
$ \Rightarrow bf = a\left( {b - f} \right)$
Now divide each side of the equation by $b - f$
$ \Rightarrow \dfrac{{bf}}{{b - f}} = \dfrac{{a\left( {b - f} \right)}}{{b - f}}$
Cancel the term and we get
$ \Rightarrow \dfrac{{bf}}{{b - f}} = a$
Thus we get the required answer.