Find the value of k for which the following system of equations has no solution:
$3x-4y+7=0$ and $kx+3y-5=0$.
Answer
635.7k+ views
Hint: In this problem, we will the condition for which a system of linear equations has no solution which is given as $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ . After applying this condition to the given system of linear equations, we can find the value of k.
Complete Step-by-Step solution:
A system of linear equations can have no solution, a unique solution or infinitely many solutions. A given system of linear equations has no solution if the equations are inconsistent.
If a system of linear equation is given as:
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 \\
\end{align}$
For the above system of linear equations to be inconsistent, it must satisfy the condition that:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ , that is the graph of the equations must be parallel to each other.
For the system of linear equations to have infinite solutions, it must satisfy the condition that $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, that is the graph of the equations must coincide.
For the system of linear equations to have a unique solution:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$
Now, the system of linear equations given to us is:
$\begin{align}
& 3x-4y+7=0 \\
& kx+3y-5=0 \\
\end{align}$
For this given system of equations to have no solution, it should satisfy:
$\dfrac{3}{k}=\dfrac{-4}{3}\ne \dfrac{7}{-5}$
So, we have:
$\begin{align}
& \dfrac{3}{k}=\dfrac{-4}{3} \\
& \Rightarrow 9=-4k \\
& \Rightarrow k=\dfrac{-9}{4} \\
\end{align}$
Also, we have:
$\begin{align}
& \dfrac{3}{k}\ne \dfrac{-7}{5} \\
& \Rightarrow 15\ne -7k \\
& \Rightarrow k\ne \dfrac{-15}{7} \\
\end{align}$
Hence, the value of k is equal to $\dfrac{-9}{4}$ given equation will have no solution.
Note: Students should note here that for a system of linear equations to be inconsistent, the graph of the equations must be parallel to each other. So, the ratio of coefficients of x and y must be equal and it should not be equal to the ratio of the constant term.
Complete Step-by-Step solution:
A system of linear equations can have no solution, a unique solution or infinitely many solutions. A given system of linear equations has no solution if the equations are inconsistent.
If a system of linear equation is given as:
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 \\
\end{align}$
For the above system of linear equations to be inconsistent, it must satisfy the condition that:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ , that is the graph of the equations must be parallel to each other.
For the system of linear equations to have infinite solutions, it must satisfy the condition that $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, that is the graph of the equations must coincide.
For the system of linear equations to have a unique solution:
$\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$
Now, the system of linear equations given to us is:
$\begin{align}
& 3x-4y+7=0 \\
& kx+3y-5=0 \\
\end{align}$
For this given system of equations to have no solution, it should satisfy:
$\dfrac{3}{k}=\dfrac{-4}{3}\ne \dfrac{7}{-5}$
So, we have:
$\begin{align}
& \dfrac{3}{k}=\dfrac{-4}{3} \\
& \Rightarrow 9=-4k \\
& \Rightarrow k=\dfrac{-9}{4} \\
\end{align}$
Also, we have:
$\begin{align}
& \dfrac{3}{k}\ne \dfrac{-7}{5} \\
& \Rightarrow 15\ne -7k \\
& \Rightarrow k\ne \dfrac{-15}{7} \\
\end{align}$
Hence, the value of k is equal to $\dfrac{-9}{4}$ given equation will have no solution.
Note: Students should note here that for a system of linear equations to be inconsistent, the graph of the equations must be parallel to each other. So, the ratio of coefficients of x and y must be equal and it should not be equal to the ratio of the constant term.
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