To solve the system of equations
[tex]\[
\begin{array}{l}
4x - y = 12 \\
x - \frac{1}{4}y = 3
\end{array}
\][/tex]
we can use the method of substitution or elimination. Here, we'll use the elimination method to find the solution.
First, let's rewrite the equations neatly:
1. [tex]\( 4x - y = 12 \)[/tex] -- (Equation 1)
2. [tex]\( x - \frac{1}{4}y = 3 \)[/tex] -- (Equation 2)
To eliminate one of the variables, we'll first clear the fraction in Equation 2 by multiplying through by 4:
[tex]\[
4x - y = 12 \quad \text{-- (Equation 1, unchanged)}
\][/tex]
[tex]\[
4 \left(x - \frac{1}{4}y\right) = 4 \cdot 3
\][/tex]
[tex]\[
4x - y = 12 \quad \text{-- (Resulting Equation 2)}
\][/tex]
Notice that the resulting Equation 2 is actually identical to Equation 1. This tells us that the two equations are not independent; they represent the same line. Therefore, every point on this line is a solution to the system.
Given that both equations represent the same line, the system is consistent and has infinitely many solutions. Therefore, we can express the solution set as:
[tex]\[
\{(x, y) \mid 4x - y = 12\}
\][/tex]
To describe the solution set in terms of one variable, solve for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[
4x - y = 12 \implies y = 4x - 12
\][/tex]
Thus, any point [tex]\((x, 4x - 12)\)[/tex] is a solution to this system. The solution can be described as:
[tex]\[\{(x, 4x - 12) \mid x \text{ is any real number}\}\][/tex]
Since the system of equations does not have a unique solution, it does not have exactly one solution and it is consistent with dependent equations.
So, the final answer is:
The system has infinitely many solutions. The solution set is [tex]\(\{(x, 4x - 12) \mid x \text{ is any real number}\}\)[/tex].
The system is consistent.
The equations are dependent.
