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Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution. {:[-2x+3y=-1],[2x-6y=4]:} No Solutions Infinitely Many Solutions One Solution

Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.\newline−2x+3yamp;=−12x−6yamp;=4 \begin{aligned} -2 x+3 y & =-1 \\ 2 x-6 y & =4 \end{aligned} \newlineNo Solutions\newlineInfinitely Many Solutions\newlineOne Solution

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Q. Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.\newline−2x+3y=−12x−6y=4 \begin{aligned} -2 x+3 y & =-1 \\ 2 x-6 y & =4 \end{aligned} \newlineNo Solutions\newlineInfinitely Many Solutions\newlineOne Solution
  1. Given Equations: We are given the system of equations:\newline−2x+3y=−1-2x + 3y = -1\newline2x−6y=42x - 6y = 4\newlineFirst, we will look for any obvious contradictions or multiples that would indicate no solutions or infinitely many solutions.
  2. Multiplying First Equation: Let's multiply the first equation by 22 to see if it matches the second equation in any way:\newline2(−2x+3y)=2(−1)2(-2x + 3y) = 2(-1)\newline−4x+6y=−2-4x + 6y = -2\newlineNow we compare this new equation with the second equation given in the problem.
  3. Comparison with Second Equation: Comparing the new equation −4x+6y=−2-4x + 6y = -2 with the second original equation 2x−6y=42x - 6y = 4, we notice that the coefficients of yy are opposites of each other, and the coefficients of xx are also opposites. However, the constants on the right side of the equations are not multiples of each other.
  4. Conclusion: Since the coefficients of xx and yy are proportional but the constants are not, this indicates that the two lines represented by these equations are parallel and do not intersect. Therefore, the system of equations has no solutions.

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