Question
In the following system of equation determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:
$x - 3y = 3$
$3x - 9y = 2$
✓
Answer
The given system of equations may be written as,
$x - 3y - 3 = 0$
$3x - 9y - 2 = 0$
The given system of equations is of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 1, b_1 = -3, c_1 = -3$
And $a_2 = 3, b_2= -9, c_2 = -2$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-9}=\frac{1}{3}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-2}=\frac{3}{2}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given system of equation has no solutions.
$x - 3y - 3 = 0$
$3x - 9y - 2 = 0$
The given system of equations is of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 1, b_1 = -3, c_1 = -3$
And $a_2 = 3, b_2= -9, c_2 = -2$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-9}=\frac{1}{3}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-2}=\frac{3}{2}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given system of equation has no solutions.
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