On the Guided Practice tab, a system of linear equations in two variables is generated for you. Drag the coefficients, variables, and constants from the equations to the matrices. The variables should be placed in the variable matrix, which is the second matrix in the matrix equation. The first matrix is the coefficient matrix. The coefficients from an equation should be placed within the same row of the coefficients matrix, so that the first coefficient in the matrix is the coefficient that is multiplied by the first variable in the variable matrix… …and the second coefficient in the coefficients matrix is the coefficient that is multiplied by the second variable in the variable matrix. The third matrix is the constant matrix. The constants should be placed in the constant matrix in the same row as the coefficients from that equation. Click “Calculate” to express the matrix in reduced row echelon form. In this form, the type of system can be determined by inspection. Select the type of linear system. The resulting matrix of a consistent independent linear system, which has one solution, will have a coefficient matrix with ones along its diagonal from the top-left to the bottom-right and zeros in all other places. Such a matrix is in reduced row echelon form, and the solution can be determined by translating the matrix back into equations. The resulting matrix of a consistent dependent linear system, which has infinitely many solutions, will have all zeros on the lowest row of the coefficient matrix, and a zero in the lowest row of the constant matrix. The upper row of the matrix of a consistent, dependent system can be translated back into an equation, which will describe the infinitely many solutions. The resulting matrix of an inconsistent linear system, which has no solutions, will have all zeros on the lowest row of the coefficient matrix, and a non-zero number in the lowest row of the constant matrix. The final row of the matrix of an inconsistent system can be translated back into an equation that will be mathematically false, thereby indicating that the matrix has no solutions. If the system is consistent and independent, drag and drop the constants into the boxes. These values make up the ordered pair that is the solution. Click “Try Another” to try another example. On the Workspace tab, you can enter your own linear equations. Click on a box to use the on-screen keypad to enter coefficients and constants… …use the dropdown menus to enter variables. Then click “Check Equations”. Drag the coefficients, variables, and constants from the equations to the matrices. Click “Calculate” to solve the linear system. The virtual manipulative will determine what type of system you have entered. If the system is consistent and independent, the virtual manipulative will also give you the constants of the unique solution. Click “Try Another” to try another example.