Question
Find the value of k for which the system has (i) a unique solution, and (ii) no solution:
$kx + 2y = 5$
$3x + y = 1$
✓
Answer
Given,
$kx + 2y = 5$
$3x + y = 1$
To find: To determine for what value of k the system of equation has:
Unique solution.
No solution.
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For Unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Here,
$\frac{\text{k}}{3}\neq\frac{2}{1}$
$\text{k}\neq6$
Hence for $\text{k}\neq6$ the system of equation has unique solution.
For no solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_1}$
$\frac{\text{k}}{3}=\frac{2}{1}\neq\frac{5}{1}$
$\text{k}=6$
Hence for k = 6 the system of equation has no solution.
$kx + 2y = 5$
$3x + y = 1$
To find: To determine for what value of k the system of equation has:
Unique solution.
No solution.
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For Unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Here,
$\frac{\text{k}}{3}\neq\frac{2}{1}$
$\text{k}\neq6$
Hence for $\text{k}\neq6$ the system of equation has unique solution.
For no solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_1}$
$\frac{\text{k}}{3}=\frac{2}{1}\neq\frac{5}{1}$
$\text{k}=6$
Hence for k = 6 the system of equation has no solution.
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