Answer:
Step-by-step explanation:
Equations [tex]a_1x+b_1y+c_1=0\,\,,\,\,a_2x+b_2y+c_2=0[/tex] have
1.unique solution if [tex]\frac{a_1}{a_2}\neq \frac{b_1}{b_2}[/tex]
2. infinite solutions if [tex]\frac{a_1}{a_2}= \frac{b_1}{b_2}=\frac{c_1}{c_2}[/tex]
3. no solution if [tex]\frac{a_1}{a_2}= \frac{b_1}{b_2}\neq \frac{c_1}{c_2}[/tex]
We can write equations: [tex]y=2x-5\,\,,\,\,-8x-4y=-20[/tex] as follows:
[tex]2x-y-5=0\,\,,\,\,-8x-4y+20=0[/tex]
On comparing these equations with standard equations [tex]a_1x+b_1y+c_1=0\,\,,\,\,a_2x+b_2y+c_2=0[/tex] , we get [tex]a_1=2\,,b_1=-1\,,\,c_1=-5\,,a_2=-8\,,b_2=-4\,,c_2=20[/tex]
such that
[tex]\frac{a_1}{a_2}=\frac{2}{-8}=\frac{-1}{4}\\\frac{b_1}{b_2}=\frac{-1}{-4}=\frac{1}{4}\\\frac{c_1}{c_2}=\frac{-5}{20}=\frac{-1}{4}[/tex]
Here, [tex]\frac{-1}{4}=\frac{a_1}{a_2}\neq \frac{b_1}{b_2}=\frac{1}{4}[/tex]
So, according to the facts explained above,
The given linear system has a unique solution i.e only one solution.
