[6x+2y=3], [6x+y=3] Consider the given system of equations. How many (x,y) solutions does this system have? Choose 1 answer: (A) No solutions (B) Exactly one solution (C) Infinitely many solutions (D) None of the above

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[6x+2y=3], [6x+y=3] Consider the given system of equations. How many  (x,y) solutions does this system have? Choose 1 answer: (A) No solutions (B) Exactly one solution (C) Infinitely many solutions (D) None of the above

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Analyze Equations: Let's analyze the system of equations: $$ \begin{cases}  6x + 2y = 3 \ 6x + y = 3  \end{cases} $$ We will subtract the second equation from the first to eliminate the variable x and find the value of y.Subtract Equations: Subtracting the second equation from the first: $$ (6x + 2y) - (6x + y) = 3 - 3 $$ $$ 6x + 2y - 6x - y = 0 $$ $$ 2y - y = 0 $$ $$ y = 0 $$Find y: Now that we have the value of y, we can substitute it back into one of the original equations to find the value of x. Let's use the second equation: $$ 6x + y = 3 $$ $$ 6x + 0 = 3 $$ $$ 6x = 3 $$ $$ x = \frac{3}{6} $$ $$ x = \frac{1}{2} $$Substitute y into Equation: We have found a single solution for the system of equations: $ x = \frac{1}{2} $ and $ y = 0 $. This means that there is exactly one solution to the system.

$6x+2y=3$ $\newline$ $6x+y=3$ $\newline$ Consider the given system of equations. How many $(x,y)$ solutions does this system have? $\newline$ Choose $1$ answer: $\newline$ (A) No solutions $\newline$ (B) Exactly one solution $\newline$ (C) Infinitely many solutions $\newline$ (D) None of the above

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