Question
Are the following pair of linear equations consistent? Justify your answer:
2ax + by = a and 4ax + 2by - 2a = 0; a, b ≠ 0
2ax + by = a and 4ax + 2by - 2a = 0; a, b ≠ 0
✓
Answer
For consistent system of linear equations a, $\text{b}\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
2ax + by = a and 4ax + 2by – 2a = 0
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{2\text{a}}{4\text{a}}=\frac{1}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{\text{b}}{2\text{b}}=\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{\text{a}}{2\text{a}}=\frac{1}{2}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of linear equations si consistent and has infinitely many solutions.
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
2ax + by = a and 4ax + 2by – 2a = 0
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{2\text{a}}{4\text{a}}=\frac{1}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{\text{b}}{2\text{b}}=\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{\text{a}}{2\text{a}}=\frac{1}{2}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of linear equations si consistent and has infinitely many solutions.
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