6. \( \left\{\begin{array}{l}x-4 y=0 \\ x=y+3\end{array}\right. \)

To solve the system of equations \( \left\{\begin{array}{l}x-4y=0 \\ x=y+3\end{array}\right. \), we can start by substituting the second equation into the first. From the first equation, we can express \( x \) in terms of \( y \): \( x = 4y \). Now, substituting \( 4y \) for \( x \) in the second equation gives us \( 4y = y + 3 \). Simplifying this leads to \( 3y = 3 \), hence \( y = 1 \). Plugging \( y = 1 \) back into \( x = 4y \) yields \( x = 4(1) = 4 \). Thus, the solution is \( (x, y) = (4, 1) \). To visualize this solution, think of each equation as a line on a graph. The first line \( x = 4y \) will pass through points like \( (0, 0) \) and \( (4, 1) \), while the second line \( x = y + 3 \) will start at \( (3, 0) \) and slope upwards. The intersection point \( (4, 1) \) is where the two lines meet, which represents the solution to the system. Drawing this out could help you see how linear equations interact!