Question

a. Find a value of \(n\) so that the linear system has infinitely many solutions. b. Find a value of \(n\) so that the linear system has no solution. c. Graph both results. $$ \begin{aligned}&x-y=3\\\&4 x-4 y=n\end{aligned} $$

Short answer

When \(n = 12\), the system of equations has infinitely many solutions, and when \(n = 10\), the system has no solution.

Step-by-step solution

Find a value of \(n\) where the system has infinitely many solutions.

For the system of equations to have infinitely many solutions, the two equations must be multiples of each other. The second equation can be simplified to \(x - y = n/4\). Now, for the system to have infinitely many solutions, \(n/4 = 3\), which implies \(n = 12\). Therefore, when \(n = 12\), the system has infinitely many solutions.

Find a value of \(n\) where the system has no solution

For the system of equations to have no solution, the lines represented by the equations must be parallel. This occurs when the coefficients of \(x\) and \(y\) in the two equations are the same, but the constants are different. The coefficients of \(x\) and \(y\) are already the same in the two equations. By choosing a \(n\) such that \(n/4\) is different from 3, we can ensure that the system has no solution. For example, when \(n = 10\), the system has no solution.

Graph both results

Plot the equations \(x - y = 3\), \(x - y = n/4\) where \(n = 12\) and \(n = 10\). The graph when \(n = 12\) will show two coincident lines showing infinitely many solutions, while the graph when \(n = 10\) will show two parallel lines showing no solution.