Question

Show that the following system of equations has a unique solution :$3x+5y=12,5x+3y=4$.

Answer 1

Hint: Assume the two equations as equation (i) and (ii). Now, write these equations in the form ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$. Here, ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ are the coefficients of $x,y$ and constant term respectively of equation (i) and similarly ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ are the coefficients of $x,y$ and constant term respectively of equation (ii). Now find the ratios $\dfrac{{{a}_{1}}}{{{a}_{2}}},\dfrac{{{b}_{1}}}{{{b}_{2}}}$ and $\dfrac{{{c}_{1}}}{{{c}_{2}}}$ and consider the cases, if $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$, then the system has unique solution, if $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$, then the system has no solution and if $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, then the system has infinitely many solutions. Find which of these conditions is satisfied by the given system of equations to get the answer.

Complete step-by-step solution:
Here, we have been provided with a system of linear equations $3x+5y=12,5x+3y=4$ and we have to show that they have a unique solution.
Now, first let us write these equations in standard form as $Ax+By+C=0$. So, we get the two equations as,
$\begin{align}
  & 3x+5y-12=0\ldots \ldots \ldots \left( i \right) \\
 & 5x+3y-4=0\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
Considering equation (i), we have,
${{a}_{1}}=$coefficient of $x=3$
${{b}_{1}}=$coefficient of $y=5$
${{c}_{1}}=$coefficient of $-12$
Considering equation (ii), we have,
${{a}_{2}}=$coefficient of $x=5$
${{b}_{2}}=$coefficient of $y=3$
${{c}_{2}}=$coefficient of $-4$
Now, to determine the number of solutions of two equations given as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$, the following three cases arise :
1. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then unique solution.
2. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$then no solution.
3. If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$then infinitely many solutions.
So, let us check which of these cases is satisfied by the equations provided to us. So, we have,
$\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{3}{5},\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{5}{3},\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-12}{-4}=3$
Clearly, we can see that $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ and therefore case 1 is satisfied. Therefore , the given system of equations has a unique solution.
Hence, proved.

Note: One may note that there can be another method also to check our answer. We can apply the graphical method for the same proof. What we can do is we will draw the graphs of the given lines represented by linear equations by plotting points. Now, we will check three cases, if the lines cut at a point then they have a unique solution if the lines are parallel but not coincident then they have no solution and if the lines are coincident then they have infinitely many solutions.