Question
Do the following pair of linear equations have no solution? Justify your answer:$3\text{x}+\text{y}-3=0$, $2\text{x}+\frac{2}{3}\text{y}=2$

Answer

The system of linear equations has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$3\text{x}+\text{y}-3=0$ and $2\text{x}+\frac{2}{3}\text{y}=2$
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{3}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{1}{\frac{2}{3}}$, $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-2}=\frac{3}{2}$
So, the given system of linear equations does not satisfy $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$.

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