Consider reducing a matrix to row-echelon form. Pretend we have a pivot (the entry we want to set to 1) equal to 3 and another entry in the same column with an entry equal to 5. We can make the pivot equal to 1 by multiplying every entry in its row by 1/3. But the result is that the problem with the other rows explode into a mess of fractions. Most people accept that and continue working with fractions throughout the entire process. A simple way to prevent the fraction problem is to replace the row containing the pivot with itself multiplied by 3 and subtract it from the row with a 5 in its column, multiplied by 2. This gives 10-9=1 for the pivot and it is a perfectly legal operation as long as we set the pivot row as a linear combination of itself and any other row.