How many Solutions does this system have? (1 point)2x+y=36x=9-3yA)oneB)noneC)infiniteD)two

The given system of equation that is [tex]2x+y=3[/tex] and [tex]6x=9-3y[/tex] has infinite number of solutions.Option -C.Solution:Need to determine number of solution given system of equation has.[tex]\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}[/tex]Let us first bring the equation in standard form for comparison[tex]\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}[/tex][tex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}[/tex]To check how many solutions are there for system of equations [tex]a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0[/tex], we need to compare ratios of [tex]\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}[/tex]In our case,  [tex]a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3[/tex][tex]a_{2}  = 6, b_{2} = 3,\text{ and }c_{2} = -9[/tex][tex]\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}[/tex]As [tex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/tex], so given system of equations have infinite number of solutions.Hence, we can conclude that system has infinite number of solutions.