 Questions & Answers    Question Answers

# 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 Verified
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.
Bookmark added to your notes.
View Notes
Coordinate Geometry For Class 10  Coordinate System  How to Solve the System of Linear Equations in Two Variables or Three Variables?  Differential Equations For Class 12  Coordinate Geometry  Number System For Class 9  Maths Coordinate Geometry Formulas  Two Dimensional Coordinate Geometry  Determine The Order of Matrix  CBSE Class 9 Maths Chapter 3 - Coordinate Geometry Formulas  