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How do you isolate a in the equation $|a+\dfrac{b}{c}|(d-e)=f$?

Last updated date: 15th Jul 2024
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Hint: We are given an equation $|a+\dfrac{b}{c}|(d-e)=f$ and we are asked to isolate a, we start our solution by understanding what does isolation of “a” mean, then we will use BODMAS, then we will learn how ‘||’ mod function behave, once we have that we will use that $|x|=c$ mean $x=\pm c$ after expanding modulus we will use an algebraic tool to expand and simplify.

Complete step-by-step solution:
We are given a function as $|a+\dfrac{b}{c}|(d-e)=f$. We have to isolate “a”, we will understand what does the isolation of “a” means as in our function we have many other forms with the term “a”, isolation of a mean we have to separate the “a” from the rest of the other term given with “a”. So, to do so we will firstly open the bracket. Simplify it.
Now, we have
As we can see that $|a+\dfrac{b}{c}|\,\,\operatorname{and}\,\,(d-e)$ and being multiplied, so we divide both sides by $d-e$ to separate from the left side. So, we divide $d-e$ on both sides,
$\Rightarrow \dfrac{|a+\dfrac{b}{c}|(d-e)}{d-e}=\dfrac{f}{d-e}$
Now we will learn that how mod of n behaves
For $|x|=c$, mod of x always gives positive value it means for $x=c$ or $-c$ it gives $|x|=c(>0)$only.
$|x|=c\,\,\,\,\,\,\,\,\,\Rightarrow x=\pm c$
Hence by using this in equation (i), we get
$a+\dfrac{b}{c}=\pm \dfrac{f}{d-e}$
Now we will need just to separate $\dfrac{b}{c}$ from the right. So we will subtract $\dfrac{b}{c}$ on both sides.
$\Rightarrow a+\dfrac{b}{c}-\dfrac{b}{c}=\pm \dfrac{f}{(d-e)}-\dfrac{b}{c}$
Simplifying we get,
$a=\pm \dfrac{f}{(d-e)}-\dfrac{b}{c}$
Hence, we isolated “a” and get that “a” in same as $a=\pm \dfrac{f}{(d-e)}-\dfrac{b}{c}$

Note: Remember that it is necessary to include all possible value then only we will reach the correct solution, if we just correct $|x|=c$ implies $x=+c$, then we will be neglecting a whole setup point, which is “a”
Also remember that we are able to divide by $d-e$ as $d-e\ne 0$so we have zero we cannot divide, we need to be clear with this.