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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

6x+2y=3 6 x+2 y=3 \newline6x+y=3 6 x+y=3 \newlineConsider the given system of equations. How many (x,y) (x, y) solutions does this system have?\newlineChoose 11 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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Full solution

Q. 6x+2y=3 6 x+2 y=3 \newline6x+y=3 6 x+y=3 \newlineConsider the given system of equations. How many (x,y) (x, y) solutions does this system have?\newlineChoose 11 answer:\newline(A) No solutions\newline(B) Exactly one solution\newline(C) Infinitely many solutions\newline(D) None of the above
  1. Analyze System of Equations: Let's analyze the system of equations:\newline{6x+2y=36x+y=3 \begin{cases} 6x + 2y = 3 \\ 6x + y = 3 \end{cases} \newlineWe will subtract the second equation from the first to eliminate the variable x and find the value of y.
  2. Subtract to Eliminate Variable: Subtracting the second equation from the first gives us:\newline(6x+2y)−(6x+y)=3−3 (6x + 2y) - (6x + y) = 3 - 3 \newline6x+2y−6x−y=0 6x + 2y - 6x - y = 0 \newline2y−y=0 2y - y = 0 \newliney=0 y = 0
  3. Find Value of x: 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:\newline6x+y=3 6x + y = 3 \newline6x+0=3 6x + 0 = 3 \newline6x=3 6x = 3 \newlinex=36 x = \frac{3}{6} \newlinex=12 x = \frac{1}{2}
  4. Final answer: We have found a single solution for the system of equations: x=12 x = \frac{1}{2} and y=0 y = 0 . This means that the system has exactly one solution.

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