How do you solve for a in $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{f}$ ?
Answer
621.6k+ views
Hint:Take out all the like terms to one side and all the alike terms to the other side. Take out all the common terms. Reduce the terms on the both sides until they cannot be reduced any further if possible. Then finally evaluate the value of the unknown variable.
Complete step by step answer: First we will start off by cross multiplying the terms in the expression.
$
\,\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{f} \\
\dfrac{{a + b}}{{ab}} = \dfrac{1}{f} \\
f(a + b) = ab \\
$
Now we will take out any common terms from both sides if possible.
$
f(a + b) = ab \\
\,\,fa + fb = ab \\
$
Now we will reduce the terms on both the sides.
$
\,\,\,fa + fb = ab \\
\,\,\,fa - ab = - fb \\
a(f - b) = - fb \\
$
Now we simplify our final answer that is evaluate the value of the variable $x$.
$\begin{array}{*{20}{c}}
{a(f - b)}& = &{ - fb} \\
a& = &{\dfrac{{ - fb}}{{(f - b)}}} \\
a& = &{\dfrac{{fb}}{{(b - f)}}}
\end{array}$
Hence, the value of $a$ is $\dfrac{{fb}}{{(b - f)}}$.
Additional Information: to cross multiply terms, you will multiply the numerator in the first fraction times the denominator in the second fraction, then you write that number down. Then you multiply the numerator of the second fraction times the number in the denominator of your first fraction,and then you write that number down. By Cross multiplication of fractions, we get to know if two fractions are equal or which one is greater. This is especially useful when you are working with larger fractions that you are not sure how to reduce. Cross multiplication also helps us to solve for unknown variables in fractions.
Note: While cross multiplying the terms, multiply the terms step-by-step to avoid any mistakes.After cross multiplication, take the variables to a side and integer type of terms to other side.Reduce the terms by factorisation.
Complete step by step answer: First we will start off by cross multiplying the terms in the expression.
$
\,\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{f} \\
\dfrac{{a + b}}{{ab}} = \dfrac{1}{f} \\
f(a + b) = ab \\
$
Now we will take out any common terms from both sides if possible.
$
f(a + b) = ab \\
\,\,fa + fb = ab \\
$
Now we will reduce the terms on both the sides.
$
\,\,\,fa + fb = ab \\
\,\,\,fa - ab = - fb \\
a(f - b) = - fb \\
$
Now we simplify our final answer that is evaluate the value of the variable $x$.
$\begin{array}{*{20}{c}}
{a(f - b)}& = &{ - fb} \\
a& = &{\dfrac{{ - fb}}{{(f - b)}}} \\
a& = &{\dfrac{{fb}}{{(b - f)}}}
\end{array}$
Hence, the value of $a$ is $\dfrac{{fb}}{{(b - f)}}$.
Additional Information: to cross multiply terms, you will multiply the numerator in the first fraction times the denominator in the second fraction, then you write that number down. Then you multiply the numerator of the second fraction times the number in the denominator of your first fraction,and then you write that number down. By Cross multiplication of fractions, we get to know if two fractions are equal or which one is greater. This is especially useful when you are working with larger fractions that you are not sure how to reduce. Cross multiplication also helps us to solve for unknown variables in fractions.
Note: While cross multiplying the terms, multiply the terms step-by-step to avoid any mistakes.After cross multiplication, take the variables to a side and integer type of terms to other side.Reduce the terms by factorisation.
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