Solving a 'Harvard' University Entrance Exam Question

[00:00:00]Hello everyone. You're all welcome.

[00:00:03]Today we have a very interesting algebra math problem.

[00:00:07]Here is a + 2 a b + b is equal to 22. So here we'll try to find out the value of a b. All possible solution of this interesting algebra problem.

[00:00:19]So let's start our solution. First of all here we'll consider a b as a positive integers. They are a and b will belongs to the set of positive integers.

[00:00:30]So first of all here we'll multiply both sides of this equation with two. So this expression will become it will become two times a + 2 a b + b is equal to two times 20 two.

[00:00:49]Let's multiply two inside the parenthesis. So this will become 2 a + 2 * 2 is 4 or a b + 2 b is equal to this is 40 four.

[00:01:05]Here we can write this 4 a b as this is simply 2 a times 2 b.

[00:01:12]And there is 2 a common in both the terms. So we'll take out 2 a common.

[00:01:18]It will become two taking 2 a common this will become 1 + 2 b + 2 b is equal to 40 four.

[00:01:32]Now here there is 1 + 2 b and there's only 2 b.

[00:01:36]So here we'll try to make this term as a common term in both the terms. So therefore here we will need positive one.

[00:01:44]So that's why here we will add one to both sides of this equation.

[00:01:49]So this will become this is 2 a 1 + 2b + And here this will become 2b + 1 is equal to 40 5.

[00:02:10]And here 1 + 2b and 2b + 1 both are the same.

[00:02:14]So we will take out 2b + 1 as a common factor. This will become taking 2b + 1 common. Here only 2a is left.

[00:02:25]Plus here only one is left is equal to 40 5.

[00:02:32]And we can also write this left-hand side as 2a + 1 * 2b + 1 is equal to 40 5.

[00:02:46]And if we substitute a and b is any positive integer, so this value will be always gives him greater than or equal to three.

[00:02:56]Similarly, for any positive integer value of b, this expression will be always greater or equal to three.

[00:03:03]So here we will factorize this 45 to only those number which are greater or equal to three.

[00:03:10]So therefore here we can write this 45 as this is 3 * 15 or 15 * 3.

[00:03:23]And we can also write this 45 as this is 5 * 9, which is 45.

[00:03:29]Or 9 * 5, which is also 45.

[00:03:35]So here we have four possible cases.

[00:03:38]So in the first case we will take this expression equal to 3 * 15 instead of 45.

[00:03:45]So our first case will become 2a + 1 * 2 b + 1 Here we'll take 45 as 3 * 15.

[00:04:01]And we'll take this expression equal to 3 and this expression equal to 15.

[00:04:06]So, from here we'll get 2 a + 1 = 3.

[00:04:12]And 2 b + 1 = 15.

[00:04:18]So, let's solve both equations for the value of a and b. So, first we will solve this one equation. This is 2 a.

[00:04:25]Here we'll take this one to the right-hand side, so it become 3 - 1 is 2.

[00:04:29]And dividing both sides by 2, this gives him a is equal to 1.

[00:04:35]Now, we'll solve this one equation, so this is 2 b.

[00:04:38]Here 15 - 1 is 14.

[00:04:41]And dividing both sides by 2, here 14 by 2, this is simply 7.

[00:04:46]So, this is our first real solution.

[00:04:48]Here we'll consider our second case.

[00:04:51]So, in our second case we will take this expression equal to this one number, 15 * 3.

[00:05:01]So, this will become 2 a + 1 * 2 b + 1 is equal to 15 * 3.

[00:05:14]Here we'll take this expression equal to 15 and this expression equal to 3.

[00:05:19]So, this will become this is 2 a + 1 is equal to 15.

[00:05:25]And 2 b + 1 is equal to 3.

[00:05:32]We'll take this one to the right-hand side, so it become 2 a is equal to 15 - 1 is 14. And dividing both sides by 2, this gives him a is equal to 7.

[00:05:43]We'll solve this one equation, so this is 2 b.

[00:05:47]3 minus 1 is 2 and dividing both sides by 2 This gives me b is equal to 1.

[00:05:54]So, this is the second real solution.

[00:05:57]And we'll take our third case. Let us here we'll take this expression equal to 5 * 9.

[00:06:05]So, our third case will become 2a + 1 * 2b + 1 is equal to 5 * 9.

[00:06:22]And here we'll take 2a + 1 is equal to 5 and 2b + 1 is equal to 9.

[00:06:35]And let's solve both the equations for the value of a and b.

[00:06:38]So, we can this is 2a.

[00:06:41]5 minus 1 is 4.

[00:06:43]Dividing both sides by 2 a is equal to 4 by 2 is simply 2.

[00:06:49]This is 2b. 9 minus 1 is 8.

[00:06:53]And dividing both sides by 2 b is equal to This is 4.

[00:06:59]We'll come to our fourth case.

[00:07:01]So, in our fourth case here we'll take this expression equal to 9 * 5.

[00:07:10]So, our fourth case will become this is 2a + 1 * 2b + 1 is equal to 9 * 5.

[00:07:23]So, this is 2a + 1 is equal to 9 and 2b + 1 is equal to 5.

[00:07:34]So, this is 2a. 9 minus 1 is 8. So, it become 8. Dividing both sides by 2 this will become A is equal to four.

[00:07:44]And this is 2B 5 - 1 is 4.

[00:07:49]Dividing both sides by two 4 by 2 is simply two.

[00:07:54]So, this is our fourth solution.

[00:07:56]So, finally here we have four possible solutions.

[00:08:00]So, these solutions will become A {comma} B that is equal to Our first solution is 1 7 and second solution is 7 1 third solution is 2 4 and the fourth solution is 4 2.

[00:08:26]So, here we have these four possible solutions of this interesting algebra math problem.

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[00:08:43]Thanks for watching.