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How many solutions does the system have? {[6x-y=-1],[6x+y=-1]:} Choose 1 answer: (A) Exactly one solution (B) No solutions (c) Infinitely many solutions

How many solutions does the system have?\newline{6x−y=−16x+y=−1 \left\{\begin{array}{l} 6 x-y=-1 \\ 6 x+y=-1 \end{array}\right. \newlineChoose 11 answer:\newline(A) Exactly one solution\newline(B) No solutions\newline(C) Infinitely many solutions

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Q. How many solutions does the system have?\newline{6x−y=−16x+y=−1 \left\{\begin{array}{l} 6 x-y=-1 \\ 6 x+y=-1 \end{array}\right. \newlineChoose 11 answer:\newline(A) Exactly one solution\newline(B) No solutions\newline(C) Infinitely many solutions
  1. Write equations: Write down the system of equations.\newlineWe have the following system of equations:\newline6x−y=−16x - y = -1\newline6x+y=−16x + y = -1
  2. Compare coefficients: Compare the two equations.\newlineBoth equations have the same coefficient for xx, which is 66. This means they have the same slope.
  3. Differences in equations: Look for differences in the equations.\newlineThe equations differ only by the sign of yy. The first equation has −y-y, and the second has +y+y.
  4. Eliminate y: Add the two equations together to eliminate y.\newline(6x−y)+(6x+y)=−1+(−1)(6x - y) + (6x + y) = -1 + (-1)\newline12x=−212x = -2
  5. Solve for x: Solve for x.\newlineDivide both sides by 1212 to isolate x:\newlinex=−212x = \frac{-2}{12}\newlinex=−16x = \frac{-1}{6}
  6. Substitute xx into equation: Substitute xx back into one of the original equations to find yy.\newlineUsing the first equation:\newline6(−16)−y=−16(-\frac{1}{6}) - y = -1\newline−1−y=−1-1 - y = -1\newliney=0y = 0
  7. Check solution in second equation: Check the solution in the second equation. \newline6(−16)+y=−16(-\frac{1}{6}) + y = -1\newline−1+y=−1-1 + y = -1\newliney=0y = 0
  8. Conclude number of solutions: Conclude the number of solutions.\newlineSince we found a single value for xx and yy that satisfies both equations, the system has exactly one solution.

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