How To Really Solve This Tricky Algebra Problem (VIII) - On the left-hand side, two equations are hand-written: xy + z = 1000; x + yz = 1001; Given these equations, on the right-hand side, the following questions are written down: x = ??; y = ??; z = ?? Below these questions, the author writes that x, y, and z belong to the set of all integers.

Welcome to the eighth entry in the tricky algebra problem series. It has been a while since I wrote about algebra problems. So, here we are.

The problem in this entry is tricky, but not very challenging like the previous one. All it requires is a bit of application and algebraic reasoning. You are presented with the following two equations:

xy + z = 1000

x + yz = 1001

where x, y, and z are all integers.

Given these initial conditions, you are required to solve for all three unknowns (x, y, and z). We essentially have three unknowns and two equations here. So, you might think that a solution is out of reach. But if you apply algebraic reasoning, you can find a way.

Do you think you can solve it?

Spoiler Alert

I will be explicitly discussing the solution to this problem beyond this point. So, I suggest that you pause reading this essay at this point and go ahead with your solution if your goal is to solve the problem independently.

Once you are done with your attempt, you may come back and continue reading to compare our approaches.

This essay is supported by Generatebg

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Setting Up the Tricky Algebra Problem

We will begin by labelling our equations for ease of reference later:

How To Really Solve This Tricky Algebra Problem (VIII) —xy + z = 1000 → A; x + yz = 1001 → B.
Equations A and B — Math illustrated by the author

From here, the next step is not very difficult to ascertain. If you spend a moment, you would realise that the equation A can be plugged into the right-hand side of equation B as follows:

How To Really Solve This Tricky Algebra Problem (VIII) —x + yz = 1001 = 1000 + 1; But xy + z = 1000; Therefore, x + yz = xy + z + 1
Math illustrated by the author

We can simplify the above equation further to obtain a third equation (C):

How To Really Solve This Tricky Algebra Problem (VIII) — x + yz = xy + z + 1; Subtracting (xy + z) from both sides: x + yz − xy − z = 1; → x − xy + yz − z = 1; Factoring common terms out: x(1 − y) + z(y − 1) = 1 → x(1 − y) − z(1 − y) = 1 → (x − z)(1 − y) = 1 → Equation C
Deriving Equation C — Math illustrated by the author

The new equation (C) we have here is the key to solving our tricky little problem.

Solving the Tricky Algebra Problem

By definition, the product of the two parentheses in equation C HAS to equal 1. Since x, y, and z are all integers, the results of the operations inside these parentheses should also yield integers. The question then becomes:

Which two integers multiplied with each other yield 1?

There are only two possible answers to this question:

I) 1*1 = 1 → [Case-1]

II) (-1)*(-1) = 1 → [Case-2]

Let us consider the first case for starters and see what solutions we get:

Solutions for Case 1

As per the assertion of case 1, we have the following two equations (D and E):

How To Really Solve This Tricky Algebra Problem (VIII) —x − z = 1 → D; 1 − y = 1 → E.
Equations D and E — Math illustrated by the author

Straight away from equation E, we deduce that y = 0. When we plug this result into equation A, we get z = 1000.

Finally, when we plug y = 0 in equation B, we get x = 1001, which also satisfies equation D. So, our final solutions for case-1 are as follows:

How To Really Solve This Tricky Algebra Problem (VIII) — x = 1001; y = 0; z = 1000.
Solutions for Case-1 — Illustration created by the author

Solutions for Case 2

As per the assertion of case 2, we have the following two equations (F and G):

xy + z = 1000; Plugging-in y=2 → 2x + z = 1000 → H — x − z = − 1 → F; 1 − y = − 1 → G
Equations F and G — Math illustrated by the author

From equation G, we can deduce that y = 2. When we plug this result into equation A, we get a new equation (H):

How To Really Solve This Tricky Algebra Problem (VIII) — xy + z = 1000; Plugging-in y=2 → 2x + z = 1000 → H
Equation H — Math illustrated by the author

When we add equations-F and -H, we get ‘x’ as follows:

How To Really Solve This Tricky Algebra Problem (VIII) — (x − z = − 1) + (2x + z = 1000) → 3x = 999 → x = 333
Solution for x — Math illustrated by the author

When we plug (x = 333) in equation F, we get z = 334. Therefore, our final solutions for case-2 are as follows:

How To Really Solve This Tricky Algebra Problem (VIII) — x = 333; y = 2; z = 334
Solution for case-2 — Illustration created by the author

There we go. We had two cases and therefore, two sets of solutions for this problem!

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