A Clever Trick for Solving Equations from Harvard University

[00:00:00]Hello everyone and welcome back. Today we have a beautiful Harvard University admission problem that looks incredibly simple but it hides a stunning symmetry.

[00:00:10]We are given two equations x + y = 6 and x * y = 6. Our mission is to find the exact values of x and y. If you think the answers are just simple integers, think again. Hit that subscribe button and let's dive into this mathematical masterpiece. To solve this system, we will use the classic substitution method. Let's take our first equation x + y = 6 and isolate x. By subtracting y from both sides, we get a very clean expression for x in terms of y. This is our golden ticket for the next step. Now we substitute this expression for x directly into our second equation. x * y = 6. Replacing x with the quantity 6 - y, we get the quantity 6 - y * y = 6. We have successfully reduced our system of two variables down to a single variable.

[00:01:07]Let's expand the left side. Distributing the y, we get 6 y - y ^2 = 6. Moving all terms to the right side to keep the y^2 positive, we get y^2 - 6 y + 6 = 0. To solve this, we need to rearrange it into the standard quadratic form, which is a y^2 + b y + c = 0. This quadratic equation does not factor neatly into integers. So, we must call upon our ultimate weapon, the quadratic formula.

[00:01:38]For any equation in the form a y^2 + b y + c = 0, the solutions are given by y = b plus or minus the<unk> of b ^ 2 - 4 a c all / 2 a. Let's identify our coefficients here. A is 1, b is -6, and c is 6. Plugging these into the formula, we get y = -6 + or minus the<unk> of -6^ 2 - 4 * 1 * 6 all over 2 * 1. of -6 gives us pos6. Now let's simplify the discriminant which is the part under the square root. -6^2 is 36. 4 * 1 * 6 is 24. Subtracting 24 from 36 gives us exactly 12. So our equation simplifies to y = 6 + or minus the<unk> of 12 / 2.

[00:02:37]We can simplify the square<unk> of 12 further. 12 is 4 * 3 and the<unk> of 4 is 2. So the<unk> of 12 becomes 2 * the square<unk> of 3. Now we divide both terms in the numerator by 2. 6 / 2 is 3 and 2 / 2 is 1. This gives us our two beautiful exact solutions for y. So our two values for y are 3 + the<unk> of 3 and 3 - the square<unk> of 3. Remember our first equation x = 6 - y. Now we must find the corresponding x values.

[00:03:14]Let's substitute our first y value. 6 - the quantity 3 +<unk> 3 gives us 3 -<unk> 3. Therefore, x1 is 3 -<unk> 3.

[00:03:27]For the second y value, we do the exact same thing. 6 minus the quantity 3 -<unk> 3. Distributing the negative sign, the negatives cancel out, giving us 3 +<unk> 3. Notice the incredible symmetry. The x and y values are simply swapped between the two solution pairs.

[00:03:47]Therefore, our final solution pairs are x= 3 -<unk> 3 and y = 3 +<unk> 3 or x= 3 +<unk> 3 and y = 3 -<unk> 3. But as good mathematicians, we never stop here. We must verify our answers. It is time now for verification. Let's check the sum first. adding x1 and y1 3 -<unk> 3 + 3 +<unk> 3. The irrational parts cancel out perfectly leaving 3 + 3 which equals 6. The first equation is satisfied. Now let's check the product. Multiplying x1 and y1 the quantity 3 -<unk> 3 * the quantity 3 +<unk> 3. This is a classic difference of squares. a minus b * a + b = a 2 - b^ 2. So 3^ 2 -<unk> 3 2 that's 9 - 3 which equals exactly 6. Perfect.

[00:04:50]And there you have it. A seemingly simple system of equations that leads to beautiful irrational solutions. The key takeaway here is to always look for symmetry and never forget to verify your work. If you enjoyed this mathematical journey, please smash that like button, share it with your friends, and subscribe for more university admission math problems. Thanks for watching and keep solving.