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How do you solve for $u$ in $y = \dfrac{{u + 1}}{{u + 2}}?$

seo-qna
Last updated date: 22nd Jul 2024
Total views: 383.7k
Views today: 9.83k
Answer
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383.7k+ views
Hint: Firstly, multiply both sides with the denominator of right hand side, then use distributive property of multiplication to open parentheses and simplify terms, after that use algebraic operations and commutative property in order to group similar terms. Finally group all the terms consisting of $u$ at one side or at left hand side (generally) and left all terms at another side or right hand side, then divide both sides with coefficients of $u$ to get the required solution.

Complete step by step solution:
In order to solve the given algebraic equation $y = \dfrac{{u + 1}}{{u +
2}}$ for the value of $u$, we need to multiply the denominator at right hand side i.e. $(u + 2)$ to both the sides
The given algebraic equation
$ \Rightarrow y = \dfrac{{u + 1}}{{u + 2}}$
Multiplying $(u + 2)$ both sides of the equation, we will get
$
\Rightarrow y(u + 2) = \left( {\dfrac{{u + 1}}{{u + 2}}} \right) \times (u + 2) \\
\Rightarrow y(u + 2) = u + 1 \\
$
Now using distributive property of multiplication to remove the parentheses and solve the equation further,
$
\Rightarrow y \times u + y \times 2 = u + 1 \\
\Rightarrow yu + 2y = u + 1 \\
$
Subtracting $u\;{\text{and}}\;2y$ from both sides in order to send the similar terms on same sides,
$ \Rightarrow yu + 2y - u - 2y = u + 1 - u - 2y$
Using commutative property to grouping similar terms together on both sides of the equation, we will
get
$
\Rightarrow \left( {yu - u} \right) + \left( {2y - 2y} \right) = \left( {u - u} \right) + \left( {1 - 2y} \right)
\\
\Rightarrow yu - u + 0 = 0 + 1 - 2y \\
\Rightarrow yu - u = 1 - 2y \\
$
Now taking $u$ common in the left hand side, we will get
$ \Rightarrow u\left( {y - 1} \right) = 1 - 2y$
Now dividing both sides with coefficient of $u$, we will get the required solution for $u$
\[
\Rightarrow \dfrac{{u\left( {y - 1} \right)}}{{\left( {y - 1} \right)}} = \dfrac{{1 - 2y}}{{\left( {y - 1}
\right)}} \\
\Rightarrow u = \dfrac{{1 - 2y}}{{y - 1}} \\
\]
Therefore \[u = \dfrac{{1 - 2y}}{{y - 1}}\] is the desired solution for $u$ in $y = \dfrac{{u + 1}}{{u + 2}}$

Note: If we switch the terms $u\;{\text{and}}\;y$ in the resultant equation, we will get the inverse function of the given function. Because when we solve for the independent variable in the equation then the final equation eventually makes the independent variable dependent and vice versa.