How To Really Solve This Tricky Algebra Problem? (IV)

Welcome to the fourth entry in the tricky algebra problem series. This problem features a system of equations involving two unknowns: x and y, both of which are limited to the set of all real numbers. To start, you are provided with the following two equations:

1. x² + y² = 7

2. x³ + y³ = 10

Given these two equations, your task is now to figure out the solution(s) for the following equation:

x + y = ??

Difficulty Alert:

Before you begin, I’d like to mention that this problem likely involves laborious methods to solve. I say “likely” because, to solve this problem succinctly (and skip the laborious work), you’d require knowledge of certain (relatively) advanced mathematical concepts.

Spoiler Alert:

If you are excited and ready to solve this equation on your own, I recommend that you pause reading this essay now and proceed with your attempt. Once you are done with your attempt, you may choose to come back and continue reading this essay. From this point onward, I will explicitly be discussing solutions to the equation.

This essay is supported by Generatebg

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Extracting Information from the First Given Equation

Let us start by assigning a third unknown to the solution(s) we are looking for by asserting the following equation:

x + y = k

Now, let us focus on the first given equation:

x² + y² = 7

Using the binomial square formula as the basis, we could express the left-hand side of the above equation as follows

x² + y² = (x + y)² — 2xy [Because (x + y)² = x² + y² + 2xy]

When we plug this result into the originally given equation, we get the following result:

Let us keep this result (equation A) at the back of our minds for now and shift our focus towards the second given equation.

Extracting Information from the Second Given Equation

As a reminder, the second given equation is as follows;

x³ + y³ = 10

Using the binomial cube formula as the basis, we could express the left-hand side of the above equation as follows:

x³ + y³ = (x + y)³ — 3xy(x + y) [Because (x + y)³ = x³ + y³ + 3xy(x + y)]

When we plug this result into the originally given equation, we get the following result:

We now have equation B, which also gives the value of ‘xy’ in terms of the unknown ‘k’. We could combine this result with equation A to proceed with the solution.

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Deriving the Cubic Equation to Solve the Tricky Algebra Problem

From equations A and B, we have the following expression:

We now have a cubic equation of the unknown we are solving for. Let us proceed with solving this equation.

Solving the Cubic Equation to Solve the Tricky Algebra Problem

When it comes to solving such an equation, my default approach is to proceed with algebraic manipulation to see if the equation yields a simpler form. In this case, it just happens so that this is the case:

The above equation, albeit simpler than the original equation we started with, can still further be simplified as follows:

We have now arrived at three potential candidates for the solution:

1. k = x + y = 1

2. k = x + y = 4

3. k = x + y = -5

There are several ways to proceed from this point on. I will present two approaches to solve this problem: a long-winded approach that uses a simpler algebraic tool (for the novice reader), and a swift approach that uses a more advanced mathematical insight (for the more advanced reader).

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The Long-winded Approach to Solve the Tricky Algebra Problem

In this approach, we use the first given equation (x² + y² = 7) and plug in the ‘y’ values from each of the potential solution equations we came up with above. The corresponding solution tracks would pan out as follows:

At the end of this rather cumbersome process, we have arrived at three quadratic equations. The good news is that we need not solve these explicitly. All we need to do is check if the discriminant is greater than or equal to zero. If you don’t know (or recollect) what a discriminant is, here is a quick crash course:

When we apply this logic to equations C, D, and E, we obtain the following result:

Finally, we have arrived at the solution. ‘x + y = 1’ is the only possible solution to this problem. Before we wind up, let us also take a look at the swift approach to solving this problem.

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The Swift Approach to Solve the Tricky Algebra Problem

In this approach, we pick up from the three ‘k’ equations we had worked out from earlier:

1. k = x + y = 1

2. k = x + y = 4

3. k = x + y = -5

From the first given equation, we know that x² + y² = 7. Knowing this, in order to look for real solutions, we can make use of a mathematical concept known as the Arithmetic Mean — Geometric Mean Inequality (known as the AM-GM Inequality).

Using this insight, we could assert the following condition:

(x + y)² ≤ 2(x² + y²)

I am skipping the derivation for this form of the AM-GM inequality in hopes that the advanced reader out there can work it out on their own. But if you do indeed need the derivation, please mention it in the comments, and I can update the essay with an appendix later (I’m too lazy to rework the shabby hand-written derivation I have at this point).

Now, the right-hand side of the inequality equation is something we could work out from the first given equation as follows:

2(x² + y²) = 2*7 = 14

The left-hand side of the equation has three possible values:

1. For k = x + y = 1 →(x + y)² = 1² = 1 [<14].

2. For k = x + y = 4 →(x + y)² = 4² = 16 [>14].

3. For k = x + y = -5 →(x + y)² = (-5)² = 25 [>14].

As we can see, ‘k = 4’ and ‘k = -5’ do not satisfy the AM-GM inequality, and are hence not valid solutions. Therefore, once again, we arrive at the solution of ‘x + y = 1’; only this time, we were much quicker!

If you managed to solve this problem using any other interesting approach, be sure to share it with us in the comments!

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Further reading that might interest you: How To Really Solve The Monkey And The Coconuts Puzzle? and How To Really Solve i^i?

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