Question

For what values of $k$, the equations $\left( 2k-1 \right)x+y=2k+1$ and $3x+y=1$ are the same?

Answer 1

Hint: Write the given two equations in the form of $ax+by+c=0$. Assume the coefficients of $x,y$ and constant term of 1st equation as ${{a}_{1}},{{b}_{1}}$ and ${{c}_{1}}$ respectively while assume the coefficients of $x,y$ and constant term of 2nd equation as ${{a}_{2}},{{b}_{2}}$ and ${{c}_{2}}$ respectively. Now, use the formula $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ and determine the value of $k$ using the relations obtained by the formula. Check all the values of $k$ by substituting it in the equation.

Complete step-by-step solution -

It is given that, we have to choose the value of $k$ so that $\left( 2k-1 \right)x+y=2k+1$ and $3x+y=1$ should be the same. That means both the equations should be coincident and have infinitely many solutions.
The condition for two equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\text{ and }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ to have infinitely many solutions is given by: $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$.
Therefore, considering $2k-1={{a}_{1}},1={{b}_{1}},2k+1={{c}_{1}}$ and $3={{a}_{2}},1={{b}_{2}},1={{c}_{2}}$, we get,
$\dfrac{2k-1}{3}=\dfrac{1}{1}=\dfrac{2k+1}{1}$.
Now, considering each equality expression one by one, we get,
Case (i): Considering, $\dfrac{2k-1}{3}=\dfrac{1}{1}$.
$\begin{align}
  & \Rightarrow 2k-1=3 \\
 & \Rightarrow 2k=4 \\
 & \Rightarrow k=2 \\
\end{align}$
Case (ii): Considering, \[\dfrac{1}{1}=\dfrac{2k+1}{1}\].
$\begin{align}
  & \Rightarrow 2k+1=1 \\
 & \Rightarrow 2k=0 \\
 & \Rightarrow k=0 \\
\end{align}$
Case (iii): Considering, $\dfrac{2k-1}{3}=\dfrac{2k+1}{1}$.
$\begin{align}
  & \Rightarrow 2k-1=6k+3 \\
 & \Rightarrow -4=4k \\
 & \Rightarrow k=-1 \\
\end{align}$
From case (i) $k=2$, we get the two equations as:
$3x+y=5\text{ and }3x+y=1$, which are not the same. Hence the value of $k$ must not be 2.
From case (ii) $k=0$, we get the two equations as:
$x+y=1\text{ and }3x+y=1$, which are not the same. Hence the value of $k$ must not be 0.
From case (iii) $k=-1$, we get the two equations as:
$-3x+y=-1\text{ and }3x+y=1$, which are not the same. Hence the value of $k$ must not be -1.
Clearly we can see that no value of $k$ satisfies the required conditions. Hence, there is no value of.

Note: After finding the values of $k$ by the relation $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, we have to check whether they satisfy the equations or not because sometimes only certain values of $k$ satisfies the conditions and not all values. In the above solution, we can see that no value of $k$ satisfies the conditions, therefore, there is no value of $k$ for which the given two equations are the same.