 You might notice that one problem with working with systems of linear equations is we have to deal with fractions. Now for many of us fractions are fun. We love dealing with additions, subtraction, multiplication, and division of fractions. But then there are some of us, maybe only 90-95% of the population, that finds dealing with fractions a little bit tedious and possibly somewhat intimidating. While it's not generally possible to completely avoid fractions, we can at least reduce the amount of time we have to spend with them through the use of re-parameterization. So if you like adding, subtracting, multiplying, and dividing fractions, you don't really need to watch this video. On the other hand, if you want to avoid working with fractions, a careful choice of parameters can help you avoid them. We'll proceed as follows. Let P1, P2, all the way up to Pm be the pivots of our row echelon matrix. Let K be the product of all of these pivots. When we choose our free variables and set up our parameters, we'll include a factor of K in all of them. For example, suppose I want to parameterize the system whose row echelon form is given. We'll go ahead and write this as a system of equations using our variables x1, x2, x3, x4, and x5. We see that x1, x3, and x4 are leading variables, while x2 and x5 are never leading variables, so x2 and x5 will have to be parameterized. Now our pivots are 3, 2, and 5, so let's include a factor of 3 times 2 times 5, that's 30, in all of our parameters. So that means we'll let x2 be 30t, and x5 is going to be 30s. Now we'll substitute these into our equations. So our last equation tells us that 5x4 plus 2x5 is 0, so I'll substitute in my value of x5 and solve for x4. The next to last row tells me 2x3 minus x4 plus 2x5 equals 0, so I'll substitute in my values for x4 and x5 and get a value for x3. The first row gives me the equation 3x1 plus 2x2 plus x3 minus 4x4 plus 5x5 equals 0. Substituting in these values, I get my value for x1. While we can't avoid fractions forever, we can minimize the amount of time we have to work with them. So if my augmented coefficient matrix has row echelon form as shown, we can parameterize our solutions as follows. This time we'll use the variables x, y, z, and w, and the two rows of our augmented coefficient matrix correspond to the two equations, 2x plus 3y plus 3z plus w equals 5, and 7y plus 5z plus 3w equals 1. x and y are leading variables, so they'll be basic variables. z and w are never leading variables, so they'll be our free variables and we'll parameterize them. Our pivots are 2 and 7, so we'll parameterize z and w with a factor of 2 times 7 or 14. So we'll make z equals 14t and w equals 14s. From the last equation, we can substitute in our values for z and w and solve for y. Our next to last equation, we can substitute in our values for y, z and w, and solve for x. And so we have our solutions for x, y, z and w.