Answer:
Option C.
Step-by-step explanation:
Let [tex]a_1x+b_1y+c_1=0[/tex] and [tex]a_2x+b_2y+c_2=0[/tex] are two line.
If [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}[/tex], then system of equations have infinite number of solutions.
If [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}[/tex], then system of equations have no solution.
If [tex]\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}[/tex], then system of equations have unique solution.
The given equations are
[tex]2x-y=7[/tex]
[tex]y=2x+3[/tex]
These equations can be rewritten as
[tex]2x-y-7=0[/tex]
[tex]2x-y+3=0[/tex]
Here, [tex]a_1=2,b_1=-1,c_1=-7,a_2=2,b_2=-1,c_2=3[/tex].
[tex]\dfrac{a_1}{a_2}=\dfrac{2}{2}=1[/tex]
[tex]\dfrac{b_1}{b_2}=\dfrac{-1}{-1}=1[/tex]
[tex]\dfrac{c_1}{c_2}=\dfrac{-7}{3}[/tex]
Since, [tex]\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}[/tex], therefore, the system of equations have no solution.
Hence, option C is correct.
