Question

# For the following system of equations, determine the value of â€˜kâ€™ for which the given system has no solution.\begin{align} & 3x-4y+7=0 \\ & kx+3y-5=0 \\ \end{align}

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 â€˜kâ€™.

To find the value of â€˜kâ€™ 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,
$3x-4y=-7$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)
$kx+3y=5$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)

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

Also, if we compare the equation (2) with ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ we will get,
${{a}_{2}}=k$, ${{b}_{2}}=3$, ${{c}_{2}}=5$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (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,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€˜kâ€™ 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,
$\therefore \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{3}{k}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (6)

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

Consider,
L.H.S. (Left Hand Side) $=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-4}{3}$ â€¦â€¦â€¦â€¦â€¦â€¦.. (9)
R.H.S. (Right Hand Side) $=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-7}{5}$â€¦â€¦â€¦â€¦â€¦â€¦ (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 â€˜kâ€™ should be equal to $-\dfrac{9}{4}$

Note: Donâ€™t use Cramerâ€™s rule in this problem as it will complicate the problem. After finding the value of â€˜kâ€™ please cross check your answer by substituting the value in $\dfrac{3}{k}=\dfrac{-4}{3}$ for perfection.