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