{:[6x+2y=3],[6x+y=3]:} Consider the given system of equations. How many (x,y) solutions does this system have? Choose 1 answer: No solutions (B) Exactly one solution C) Infinitely many solutions cord yourself and your screen at the same time (C) 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: No solutions (B) Exactly one solution C) Infinitely many solutions cord yourself and your screen at the same time (C) None of the above

Bytelearn · Accepted Answer

Analyze System of Equations: Analyze the given system of equations. We have the system: $$ \begin{cases}  6x + 2y = 3 \ 6x + y = 3  \end{cases} $$ We want to determine the number of solutions for this system.Subtract Equations to Eliminate x: Subtract the second equation from the first equation to eliminate x and solve for y. Subtracting the second equation from the first, we get: $$ (6x + 2y) - (6x + y) = 3 - 3 $$ $$ 6x + 2y - 6x - y = 0 $$ $$ 2y - y = 0 $$ $$ y = 0 $$Substitute y = 0: Substitute y = 0 into one of the original equations to solve for x. Let's substitute y = 0 into the second equation: $$ 6x + y = 3 $$ $$ 6x + 0 = 3 $$ $$ 6x = 3 $$ $$ x = \frac{3}{6} $$ $$ x = \frac{1}{2} $$Check Solution: Check the solution (x = 1/2, y = 0) in both original equations to ensure it satisfies both. First equation check: $$ 6x + 2y = 3 $$ $$ 6(\frac{1}{2}) + 2(0) = 3 $$ $$ 3 + 0 = 3 $$ $$ 3 = 3 $$ (True)  Second equation check: $$ 6x + y = 3 $$ $$ 6(\frac{1}{2}) + 0 = 3 $$ $$ 3 + 0 = 3 $$ $$ 3 = 3 $$ (True)

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