 So let's see if we can find a better way of solving linear congruences. So remember that earlier we solved the congruence 5x congruent to 1 mod 22 by listing all equations 5x equal to b, where b is congruent to 1 mod 22, and then solving one that had an integer solution. So let's think about how we can refine this approach. Rather than listing all numbers congruent to 1 mod 22, notice that all such numbers must be one more than a multiple of 22. And this suggests the following approach. Introduce a new variable y, and then find whole number solutions to 5x equals 22y plus 1. So let's see if we can find whole number solutions to 5x equals 22y plus 1. Since this is an equation, we can solve it the usual way. So if I want to solve 5x equals 22y plus 1 for x, I can't. Now remember, our goal is to find whole number solutions to this equation. And because we're dividing, we know that sometimes division produces non-whole number solutions. But sometimes it does produce whole numbers, and that's when the div at end is a multiple of the divisor. So let's split the rational expression into parts that are divisible by 5 and leftovers. So this 22y, we can split this into a part that is divisible by 5, that would be 20y, and the leftovers 2y. And the plus 1 is never going to be divisible by 5, so we'll keep that with the leftovers. But now that I have my numerator written as a sum, I could split the fraction into a sum of 2 fractions, 20y over 5, and 2y plus 1 over 5. And this first part can be reduced, leaving us with 4y plus 2y plus 1 over 5. And so what this means is that if 5x equals 22y plus 1 and x and y are both whole numbers, then x equals 4y plus 2y plus 1 over 5. But in order for x to be a whole number, the fractional part of this expression has to also be a whole number. And so I know that 2y plus 1 over 5 must be a whole number. Well, now let's think about that. Since we want 2y plus 1 over 5 to be a whole number, let's introduce another variable, z, and get z equals 2y plus 5 over 1. And since we want to know what y is, let's solve this equation for y. Again, if we want y to be a whole number, we need this fraction 5z minus 1 over 2 to also be a whole number. So again, let's split the fraction into a part that can be divided by 2 and some leftovers. So 5z, well that's really 4z plus z and the minus 1 we're stuck with. And we can split up our fraction and simplify. And that tells us that y is equal to 2z plus z minus 1 over 2, where if we want y to be a whole number, z minus 1 over 2 must be a whole number. And once again, since we want z minus 1 to be a whole number, let's introduce a new variable, w, with w equal to z minus 1 over 2, and solving for z gives us. But notice that we can find z without any sort of division. And so that as long as w is a whole number, z will be as well. And so now we can work our way backwards. As long as w is a whole number, z will be a whole number, where z minus 1 over 2 is a whole number. And that means y will be a whole number, which means that 2y plus 1 over 5 will be a whole number, and x will be a whole number, which is what we're looking for. And what this means is we can choose any whole number we want to for w. So let's pick w equals 0. If w equals 0, then we'll find z, because z is 2w plus 1, so z is equal to 1. But now that I know z is equal to 1, I can find y, because y is 2z plus z minus 1 over 2. And so y is equal to 2. But now that I know the value of y, I can find the value of x, because x equals 4y plus 2y plus 1 over 5, so x is equal to 9.