Question

Find the value of k if the following equations have infinitely many solutions
\[\begin{align}
  & (k-3)x+3y=k \\
 & kx+ky=12 \\
\end{align}\]

Answer 1

Hint: A given set of linear equations \[{{a}_{1}}x+{{b}_{1}}y={{c}_{1}},{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\]will have infinitely many solutions if\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]. Now we should compare \[{{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\] with \[(k-3)x+3y=k\]. We will get the values of \[{{a}_{1}},{{b}_{1}},{{c}_{1}}\]. Now we should compare \[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\] with \[kx+ky=12\]. We will get the values of \[{{a}_{2}},{{b}_{2}},{{c}_{2}}\]. Now we have to calculate the value of k where \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\].

Complete step-by-step answer:
Let us consider two equations as shown below.
\[\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y={{c}_{1}}......(1) \\
 & {{a}_{2}}x+{{b}_{2}}y={{c}_{2}}......(2) \\
\end{align}\]
We should know that equation (1) and equation (2) will have infinitely many solutions if \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\].
From the question, we were given the set of linear equations as shown below.
\[\begin{align}
  & (k-3)x+3y=k \\
 & kx+ky=12 \\
\end{align}\]
Let us assume
\[\begin{align}
  & (k-3)x+3y=k.........(3) \\
 & kx+ky=12..........(4) \\
\end{align}\]
Now let us compare equation (3) and equation (1), we get
\[\begin{align}
  & {{a}_{1}}=k-3......(5) \\
 & {{b}_{1}}=3..........(6) \\
 & {{c}_{1}}=k.........(7) \\
\end{align}\]
Now let us compare equation (4) and equation (2), we get
\[\begin{align}
  & {{a}_{2}}=k......(8) \\
 & {{b}_{2}}=k..........(9) \\
 & {{c}_{2}}=12.........(10) \\
\end{align}\]
From equation (5) and equation (8), we can find the value of \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\].
\[\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{k-3}{k}...(11)\]
From equation (6) and equation (9), we can find the value of \[\dfrac{{{b}_{1}}}{{{b}_{2}}}\].
\[\Rightarrow \dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{3}{k}...(12)\]
From equation (7) and equation (10), we can find the value of \[\dfrac{{{c}_{1}}}{{{c}_{2}}}\].
\[\Rightarrow \dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{k}{12}...(13)\]
Now we have to equate equation (11) and equation (12), we get
\[\begin{align}
  & \Rightarrow \dfrac{k-3}{k}=\dfrac{3}{k} \\
 & \Rightarrow k-3=3 \\
 & \Rightarrow k=6 \\
\end{align}\]
Now we have to equate equation (12) and equation (13), we get
\[\Rightarrow \dfrac{3}{k}=\dfrac{k}{12}\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow {{k}^{2}}=36 \\
 & \Rightarrow k=-6,6 \\
\end{align}\]
Now we have to equate equation (11) and equation (13), we get
\[\Rightarrow \dfrac{k-3}{k}=\dfrac{k}{12}\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow 12k-36={{k}^{2}} \\
 & \Rightarrow {{k}^{2}}-12k+36=0 \\
 & \Rightarrow {{(k-6)}^{2}}=0 \\
 & \Rightarrow k=6 \\
\end{align}\]
So, we can say that \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]if the value of k is equal to 6.

Note: Students may have a misconception that both -6 and 6 are both the values of k. But we cannot take the value of k as -6 because if k=-6 then we get\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\]. But the condition for which the set of linear equations will have infinitely many solutions if \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]. So, we cannot take the value of k to be equal to -6. Students should have a clear view to solve this problem.