Answer:
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}x-3y=15\\3x-9y=45\end{array}\right\\\\\text{Divide both sides of the second equation by 3}\\\\\dfrac{3x}{3}-\dfrac{9y}{3}=\dfrac{45}{3}\\\\x-3y=15\\\\\text{We received the same equation as the first.}\\\\\bold{CONCLUSION}\\\\\text{The system of equations has infinitely many solutions.}[/tex]
[tex]\left\{\begin{array}{ccc}a_1x+b_1y=c_1\\a_2x+b_2x=c_2\end{array}\right\\\\\text{If}\ a_1=a_2,\ b_2=b_2,\ c_1=c_2,\ \text{then the system of equations}\\\text{ has infinitely many solutions}\\\\\text{If}\ a_1=a_2,\ b_1=b_2,\ c_1\neq c_2,\ \text{then the system of equations}\\\text{has no solutions}\\\\\text{Other the system of equations has one solution.}[/tex]