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How many solutions does the system of equations below have?\newline2x+2y2z=102x + 2y - 2z = -10\newlinex2y3z=15-x - 2y - 3z = 15\newline2x+3y+2z=202x + 3y + 2z = -20\newlineChoices:\newline(A)no solution\newline(B)one solution\newline(C)infinitely many solutions

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Q. How many solutions does the system of equations below have?\newline2x+2y2z=102x + 2y - 2z = -10\newlinex2y3z=15-x - 2y - 3z = 15\newline2x+3y+2z=202x + 3y + 2z = -20\newlineChoices:\newline(A)no solution\newline(B)one solution\newline(C)infinitely many solutions
  1. Write Equations: First, let's write down the system of equations to see what we're working with:\newline11) 2x+2y2z=102x + 2y - 2z = -10\newline22) x2y3z=15-x - 2y - 3z = 15\newline33) 2x+3y+2z=202x + 3y + 2z = -20
  2. Eliminate Variable yy: We can try to simplify the system by adding equations (11) and (22) to eliminate the variable yy. So, (2x+2y2z)+(x2y3z)=10+15(2x + 2y - 2z) + (-x - 2y - 3z) = -10 + 15 This simplifies to x5z=5x - 5z = 5
  3. Correct Mistake: Now let's add equations (11) and (33) to eliminate yy again. So, (2x+2y2z)+(2x+3y+2z)=1020(2x + 2y - 2z) + (2x + 3y + 2z) = -10 - 20 This simplifies to 4x+5y=304x + 5y = -30 But wait, I made a mistake here. I should have canceled yy, not kept it. Let's correct that.

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