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:
$3x - 5y = 20$
$6x - 10y = 40$
✓
Answer
$3x - 5y = 20$
$6x - 10y = 40$
Compareit with
$ a_1 x+b_1 y+c_1=0 $
$a_2 x+b_2 y+c_2=0$
$\text { we get }$
$ a_1=3, b_1=-5, \text { and } c_1=-20 $
$a_2=6, b_2=-10 \text { and } c_2=-40$
$\frac{\text{a}_1}{\text{a}_2}=\frac{3}{6},\frac{\text{b}_1}{\text{b}_2}=\frac{5}{10},$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Simplifyingit we get
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{1}{2},$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Hence, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So both lines are coincident and overlap with each other, so it will have infinite or many solutions.
$6x - 10y = 40$
Compareit with
$ a_1 x+b_1 y+c_1=0 $
$a_2 x+b_2 y+c_2=0$
$\text { we get }$
$ a_1=3, b_1=-5, \text { and } c_1=-20 $
$a_2=6, b_2=-10 \text { and } c_2=-40$
$\frac{\text{a}_1}{\text{a}_2}=\frac{3}{6},\frac{\text{b}_1}{\text{b}_2}=\frac{5}{10},$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Simplifyingit we get
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{1}{2},$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Hence, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So both lines are coincident and overlap with each other, so it will have infinite or many solutions.
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