Question

Solve: $2\left( 3u-v \right)=5uv;2\left( u+3v \right)=5uv$

Answer 1

Hint: In the question we are given two linear equations of two variables. First we will learn the names of methods which we can use to solve these equations. Then we will use the substitution method to solve these as it is the easiest method to get our unique solution and then we will also see trivial solutions for these equations.

Complete step by step answer:
In the question we are given two equations in two variables.
To solve linear equations in two variables we have four methods
a) Graphical Method
b) Substitution Method
c) Elimination Method
d) Cross Multiplication Method.
We can use any method to solve equations.
Now we will proceed to our equation.
In the question we are given two equations in variable $u$ and$v$.
$\begin{align}
  & 2\left( 3u-v \right)=5uv \\
 & 2\left( u+3v \right)=5uv \\
\end{align}$
Simplifying these two equations
$\begin{align}
  & 6u-2v=5uv \\
 & 2u+6v=5uv \\
\end{align}$
Dividing both equations with $uv$
$\begin{align}
  & \dfrac{6u-2v}{uv}=\dfrac{5uv}{uv} \\
 & \dfrac{2u+6v}{uv}=\dfrac{5uv}{uv} \\
\end{align}$
So our new equations become
$\begin{align}
  & \dfrac{6u}{uv}-\dfrac{2v}{uv}=5 \\
 & \dfrac{2u}{uv}+\dfrac{6v}{uv}=5 \\
\end{align}$
$\begin{align}
  & \dfrac{6}{v}-\dfrac{2}{u}=5 \\
 & \dfrac{2}{v}+\dfrac{6}{u}=5 \\
\end{align}$
To solve these equations simply,
Let $\dfrac{1}{v}=x$ and $\dfrac{1}{u}=y$
Now our equations become
$\begin{align}
  & 6x-2y=5......eq\left( 1 \right) \\
 & 2x+6y=5......eq\left( 2 \right) \\
\end{align}$
We will solve these equations using substitution method
From $eq\left( 1 \right)$ we will take out value of $x$ in terms of $y$ and we will use this in $eq\left( 2 \right)$ to make it a linear equation in one variable and we will solve this using transposition method.
$\begin{align}
  & 6x-2y=5 \\
 & \Rightarrow 6x=5+2y \\
 & \Rightarrow x=\dfrac{5+2y}{6}......eq\left( 3 \right) \\
\end{align}$
Using $eq\left( 3 \right)$ in $eq\left( 2 \right)$
$\begin{align}
  & \\
 & 2\dfrac{5+2y}{6}+6y=5 \\
 & \\
\end{align}$
Now we will simplify this
$\begin{align}
  & 2\dfrac{5+2y}{6}+6y=5 \\
 & \Rightarrow \dfrac{10+4y}{6}+6y=5 \\
 & \Rightarrow \dfrac{10+4y+36y}{6}=5 \\
 & \Rightarrow 10+40y=5\times 6 \\
 & \Rightarrow 10+40y=30 \\
\end{align}$
Now this become linear equation in one variable, where variable is $y$
$10+40y=30$
We will solve this
Transpose $10$ to right hand side
$40y=30-10$
$40y=20$
Transposing $40$ to right hand side
$\begin{align}
  & y=\dfrac{20}{40} \\
 & \Rightarrow y=\dfrac{1}{2} \\
\end{align}$
Now we use value of $y$ in $eq\left( 3 \right)$
$x=\dfrac{5+2y}{6}$
 Substituting $y=\dfrac{1}{2}$
$x=\dfrac{5+2\times \dfrac{1}{2}}{6}$
Simplifying this
$\begin{align}
  & x=\dfrac{5+1}{6} \\
 & \Rightarrow x=\dfrac{6}{6} \\
 & \Rightarrow x=1 \\
\end{align}$
So we have calculated values of $x$ and $y$
$\begin{align}
  & x=1 \\
 & y=\dfrac{1}{2} \\
\end{align}$
As we know $x=\dfrac{1}{v}$
So, substituting value of $x$
$\begin{align}
  & 1=\dfrac{1}{v} \\
 & \Rightarrow v=1 \\
\end{align}$
$\therefore v=1$ Is our required value.
Now we will find value of $u$ using value of $y$
As we know
$y=\dfrac{1}{u}$
Substituting value of $y$
$\begin{align}
  & \dfrac{1}{2}=\dfrac{1}{u} \\
 & \Rightarrow u=2 \\
\end{align}$
$\therefore u=2$ Is our required value
Hence $v=1$ and $u=2$ are our required values.
There is one more solution which satisfies our given equations.
If we substitute $u=0$ and $v=0$ in these two equations
$\begin{align}
  & 6u-2v=5uv \\
 & 2u+6v=5uv \\
\end{align}$
We will get
$\begin{align}
  & 6\left( 0 \right)-2\left( 0 \right)=5\left( 0 \right)\left( 0 \right) \\
 & \Rightarrow 0-0=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
And
$\begin{align}
  & 2\left( 0 \right)+6\left( 0 \right)=5\left( 0 \right)\left( 0 \right) \\
 & \Rightarrow 0-0=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
$\therefore u=0\And v=0$ Is also a solution for these two equations.
But this solution will be considered a trivial solution because $u=0\And v=0$ is not a particular or unique solution.

Note:
 To solve linear equations in two variables we will need at least two equations. If only one equation of two variables is given then there will be infinite solutions and there will be no unique answer. In general we can say that to solve linear equations of $n$ variables we will at least need $n$ number of equations.