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{:[-6x+4y=2],[3x-2y=-1]:} Consider the system of equations. How many (x,y) solutions does this system have? Choose 1 answer: (A) No solutions (B) Exactly one solution (C) Exactly two solutions (D) Infinitely many solutions

−6x+4y=23x−2y=−1 \begin{aligned} -6 x+4 y & =2 \\ 3 x-2 y & =-1 \end{aligned} \newlineConsider the 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) Exactly two solutions\newlineD Infinitely many solutions

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Q. −6x+4y=23x−2y=−1 \begin{aligned} -6 x+4 y & =2 \\ 3 x-2 y & =-1 \end{aligned} \newlineConsider the 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) Exactly two solutions\newlineD Infinitely many solutions
  1. Multiply by 22: First, let's multiply the second equation by 22 to make the coefficients of yy the same.\newline2×(3x−2y)=2×(−1)2\times(3x - 2y) = 2\times(-1)\newline6x−4y=−26x - 4y = -2
  2. Create new system: Now we have the system: {−6x+4y=2,6x−4y=−2}\{-6x + 4y = 2, 6x - 4y = -2\}
  3. Eliminate y: Add the two equations together to eliminate y.\newline(−6x+4y)+(6x−4y)=2+(−2)(-6x + 4y) + (6x - 4y) = 2 + (-2)\newline0=00 = 0
  4. Infinitely many solutions: Since 0=00 = 0 is a true statement and we have eliminated both variables, this means the system has infinitely many solutions.

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