A Tricky Maths Problem | Algebra problem

Added: 2026-05-12

[00:00:00]This is a fun little problem. We are given two equations. a squared minus b equals 111 and b squared minus a also equals 111.

[00:00:12]We are additionally told that a is not equal to b.

[00:00:16]Our goal is to find all possible values of a and b.

[00:00:21]Pause this video, try to solve it on your own, and let me know your answer in the comments. All right, since both equations are equal to the same number, the smartest first step is to subtract them.

[00:00:34]So, let us subtract the second equation from the first equation.

[00:00:38]On the right side, 111 minus 111 becomes zero.

[00:00:43]On the left side, we get a squared minus b minus b squared minus a.

[00:00:51]Open the brackets to get this as minus b squared plus a.

[00:00:56]Put this minus b squared with this a squared to get a squared minus b squared, and the remaining is plus a minus b.

[00:01:05]Now, a squared minus b squared is none other than the difference of squares.

[00:01:11]So, it factors into a minus b multiplied by a plus b. So, take a minus b as common from both terms. This gives a minus b multiplied by a plus b plus 1 equals zero. This gives us two possible cases. The first case is a minus b equals zero, which means a equals b.

[00:01:35]But, the question clearly states that a is not equal to b. So, we reject this case completely.

[00:01:42]Now, move to the second case where a plus b plus 1 equals zero. Rearranging this gives b equals minus a minus 1.

[00:01:52]Now, substitute this value of b back into the original equation a squared minus b equals 111. Replacing b gives a squared minus of minus a minus 1 equals 111.

[00:02:08]After opening the brackets, we get a squared plus a plus 1 equals 111.

[00:02:15]Now, bring 111 to the left side to make it minus 111 in order to form a quadratic equation.

[00:02:23]This gives a squared plus a minus 110 equals zero.

[00:02:30]We now factor this quadratic. The equation factors into a plus 11 multiplied by a minus 10 equals zero.

[00:02:38]Hence, a can either be minus 11 or 10.

[00:02:42]If a equals minus 11, then substitute into b equals minus a minus 1. This gives b equals 11 minus 1, which becomes 10. So, one solution pair is minus 11 and 10. If a equals 10, then b becomes minus 10 minus 1, which is minus 11. So, the second solution pair is 10 and minus 11.

[00:03:07]Therefore, the final solution pairs of a and b are minus 11, {comma} 10, and 10, {comma} minus 11. And that's it.

[00:03:17]Now, here's a challenge for you. What happens if we also allow the case where a equals b?

[00:03:23]Try to find the values of a and b in such a situation, and write your answer in the comments. Like, share, and subscribe. So good.