Question
Find the values of k for which the system will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?
$2x + ky = 1$
$3x - 5y = 7$
✓
Answer
The given equations are,
$2x + ky = 1 ....(i)$
$3x - 5y = 7 ......(ii)$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = k, c_1 = -1$
And, $a_2 = 3, b_2 = -5$, and $c_2 = -7$
For unique solution,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{3}\neq\frac{\text{k}}{-5}$
$\Rightarrow\text{k}\neq\frac{-10}{3}$
Thus, the given system of equations has unique solution if $\text{k}\neq\frac{-10}{3}$
For no solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}=\frac{-1}{-7}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}$ and $\frac{\text{k}}{-5}=\frac{1}{7}$
$\Rightarrow\text{k}=\frac{-10}{3}$ and $\text{k}=\frac{-5}{7}$
$2x + ky = 1 ....(i)$
$3x - 5y = 7 ......(ii)$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = k, c_1 = -1$
And, $a_2 = 3, b_2 = -5$, and $c_2 = -7$
For unique solution,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{3}\neq\frac{\text{k}}{-5}$
$\Rightarrow\text{k}\neq\frac{-10}{3}$
Thus, the given system of equations has unique solution if $\text{k}\neq\frac{-10}{3}$
For no solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}=\frac{-1}{-7}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}$ and $\frac{\text{k}}{-5}=\frac{1}{7}$
$\Rightarrow\text{k}=\frac{-10}{3}$ and $\text{k}=\frac{-5}{7}$
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