Full solution

Q. Determine if the following system of equations has no solutions, infinitely many solutions or exactly one solution.

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\begin{array}{c} 3 x+y=4 \\ 12 x+4 y=16 \end{array}

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Infinitely Many Solutions

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One Solution

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No Solutions

Check Coefficients Multiples: We are given the system of equations: $\newline$ $3x + y = 4$ $\newline$ $12x + 4y = 16$ $\newline$ First, we will check if the second equation is a multiple of the first equation.
Check Constant Term: To check if the second equation is a multiple of the first, we can divide the coefficients of the second equation by the corresponding coefficients of the first equation. $\newline$ $\frac{12}{3} = 4$ and $\frac{4}{1} = 4$ $\newline$ Since both coefficients of $x$ and $y$ in the second equation are $4$ times the corresponding coefficients in the first equation, we can conclude that the second equation is indeed a multiple of the first.
Conclusion: Infinitely Many Solutions: Now, we will check if the constant term of the second equation is also a multiple of the constant term of the first equation. $\newline$ $\frac{16}{4} = 4$ $\newline$ This shows that the constant term in the second equation is also $4$ times the constant term in the first equation.
Conclusion: Infinitely Many Solutions: Now, we will check if the constant term of the second equation is also a multiple of the constant term of the first equation. $\newline$ $\frac{16}{4} = 4$ $\newline$ This shows that the constant term in the second equation is also $4$ times the constant term in the first equation.Since all terms in the second equation are multiples of the corresponding terms in the first equation, the two equations represent the same line. Therefore, the system of equations does not have a unique solution. Instead, it has infinitely many solutions, as any point that lies on the line represented by the first equation will also lie on the line represented by the second equation.