Question

For the following system of equations, determine the value of ‘k’ for which the given system has no solution.
$\begin{align}
  & 3x-4y+7=0 \\
 & kx+3y-5=0 \\
\end{align}$

Answer 1

Hint: First compare the equations with ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and \[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\] respectively and assign the values to them. After that use the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] to find the value of ‘k’.

Complete step-by-step answer:
To find the value of ‘k’ we will first rewrite the system of equations as follows,
$\begin{align}
 & 3x-4y+7=0 \\
 & kx+3y-5=0 \\
\end{align}$

By shifting the constants on the left hand side of the equations given above we will get,
$3x-4y=-7$ ………………………………. (1)
$kx+3y=5$ ………………………………. (2)

If we compare the equation (1) with ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$we will get,
${{a}_{1}}=3$, ${{b}_{1}}=-4$, ${{c}_{1}}=-7$………………………….. (3)


Also, if we compare the equation (2) with \[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\] we will get,
\[{{a}_{2}}=k\], \[{{b}_{2}}=3\], \[{{c}_{2}}=5\] ………………………….. (4)

As given in the problem the system of equations has no solution and therefore to proceed further we should know the condition given below,

Concept:
The system of equations given by ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$and\[{{a}_{2}}x+{{b}_{2}}y={{c}_{2}}\] has no solution if,
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] ………………………………… (5)
i.e. we have to first check that whether \[\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] and then we should find the value of ‘k’ by using the equation \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\].

To verify the above condition we will find all the terms separately,
Therefore we will first find \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\] by using the equation (3) and equation (4) we will get,
\[\therefore \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{3}{k}\] …………………………………………… (6)

Then we will find the value of \[\dfrac{{{b}_{1}}}{{{b}_{2}}}\] by using the equation (3) and equation (4) we will get,
\[\therefore \dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-4}{3}\] …………………………………………… (7)
Lastly we will find the value of \[\dfrac{{{c}_{1}}}{{{c}_{2}}}\] by using the equation (3) and equation (4) we will get,
\[\therefore \dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-7}{5}\] …………………………………………… (8)
Now, to check whether \[\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] first

Consider,
L.H.S. (Left Hand Side) \[=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-4}{3}\] ……………….. (9)
R.H.S. (Right Hand Side) \[=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-7}{5}\]……………… (10)
From equation (9) and equation (10) we can say that,
\[\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\]
As the above condition is satisfied therefore to satisfy the condition given in problem \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\] must be equal to\[\dfrac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore it the system has no solution then,
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\]
If we out the value of equation (3) ad equation (4) we will get,
\[\therefore \dfrac{3}{k}=\dfrac{-4}{3}\]

By cross multiplication we can write the above equation as,
\[\therefore 3\times 3=-4\times k\]
\[\therefore 9=-4\times k\]
\[\therefore k=-\dfrac{9}{4}\]
Therefore if the system of equations has no solution then the value of ‘k’ should be equal to \[-\dfrac{9}{4}\]

Note: Don’t use Cramer’s rule in this problem as it will complicate the problem. After finding the value of ‘k’ please cross check your answer by substituting the value in \[\dfrac{3}{k}=\dfrac{-4}{3}\] for perfection.