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Linear Equations in Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations in Two Variables3 Marks
Question
For what value of $\alpha,$ the system of equations will have no solution
$\alpha\text{x}+3\text{y}=\alpha-3$
$12\text{x}+\alpha\text{y}=\alpha$
 
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Answer

The given system of equations may be written as,
$\alpha\text{x}+3\text{y}-(\alpha-3)=0$
$12\text{x}+\alpha\text{y}-\alpha=0$
The given system of equations is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $\text{a}_1=\alpha,\text{b}_1=3,\text{c}_1=-(\alpha-3)$
and, $\text{a}_2=12,\text{b}_2=\alpha,\text{c}_2=-\alpha$
For no solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\alpha}{12}=\frac{3}{\alpha}\neq\frac{-(\alpha-3)}{-\alpha}$
Now,
$\frac{3}{\alpha}\neq\frac{-(\alpha-3)}{-\alpha}$
$\Rightarrow\frac{3}{\alpha}\neq\frac{\alpha-3}{\alpha}$
$\Rightarrow3\neq\alpha-3$
$\Rightarrow3+3\neq\alpha$
$\Rightarrow6\neq\alpha$
$\Rightarrow\alpha\neq6$
and $\frac{\alpha}{12}=\frac{3}{\alpha}$
$\Rightarrow\alpha^2=36$
$\Rightarrow\alpha=\pm6$
$\Rightarrow\alpha=-6$ $\big[\therefore\alpha\neq6\big]$
Hence, the given system of equations will have no solution, if $\alpha=-6$

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