Thanks to Sean and Nick from New Zealand for the suggestion! I adapted this question from a 2020 test in New Zealand.

Solve for *x*, *y*, and *z* if:

*x*^{2} – *yz* = 2*y*^{2} – *xz* = 3*z*^{2} – *xy* = 5

In fact, solve the general case for unknown constants *a*, *b*, *c*:

*x*^{2} – *yz* = *a**y*^{2} – *xz* = *b**z*^{2} – *xy* = *c*

As usual, watch the video for a solution.

Or keep reading.

.

.

“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

.

.

.

.

.

.

M

I

N

D

.

Y

O

U

R

.

D

E

C

I

S

I

O

N

S

.

P

U

Z

Z

L

E

.

.

.

.**Answer To Three Equations Quadratic Difference**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Order the equations as (i), (ii), (iii).

*x*^{2} – *yz* = *a**y*^{2} – *xz* = *b**z*^{2} – *xy* = *c*

We will eliminate the squared terms with a neat trick. Multiply (i) by *y*, (ii) by *z*, and (iii) by *x*.

*x*^{2}*y* – *y*^{2}*z* = *a**y**y*^{2}*z* – *x**z*^{2} = *b**z**x**z*^{2} – *x*^{2}*y* = *c**x*

Sum the three equations. Miraculously all the terms on the left hand side cancel! So we get:

0 = *ay* + *bz* + *cx*

0 = (*x*, *y*, *z*) · (*c*, *a*, *b*)

But there’s another way we could have eliminated the square terms! Multiply (i) by *z*, (ii) by *x*, and (iii) by *y*.

*x*^{2}*z* – *y**z*^{2} = *a**z**x**y*^{2} – *x*^{2}*z* = *b**x**y**z*^{2} – *x**y*^{2} = *c**y*

Again sum the three equations, and again all the terms on the left hand side magically cancel! So we get:

0 = *az* + *bx* + *cy*

0 = (*x*, *y*, *z*) · (*b*, *c*, *a*)

We have derived two equations showing that the vector (*x*, *y*, *z*) is orthogonal to both vectors (*c*, *a*, *b*) and (*b*, *c*, *a*). How can we find such a vector that is mutually orthogonal to two vectors? Easy, we use use the cross product! So let’s calculate the cross product of (*c*, *a*, *b*) and (*b*, *c*, *a*) to find a mutually orthogonal vector. Then we just need to find the magnitude of (*x*, *y*, *z*). So let’s calculate the cross product.

Now we multiply this by a scalar *t* to find the solution vector

(*x*, *y*, *z*)

= *t*(*a*^{2} – *bc*, *b*^{2} – *ac*, *c*^{2} – *ab*)

= (*t*(*a*^{2} – *bc*), *t*(*b*^{2} – *ac*), *t*(*c*^{2} – *ab*))

Let’s go back to the original equations.

*x*^{2} – *yz* = *a**y*^{2} – *xz* = *b**z*^{2} – *xy* = *c*

If *a* = *b* = *c* = 0, then substituting into the solution vector gives only one solution (*x*, *y*, *z*) = (0, 0, 0). (We don’t even have to solve for *t*).

So suppose not all three constants are equal to 0. For whichever constant is non-zero, we can use the corresponding equation to solve for *t*. Let’s suppose *a* is not zero, so we will use the first equation. We substitute the solution vector into the first equation

*x*^{2} – *yz* = *a*

(*t*(*a*^{2} – *bc*))^{2} – *t*(*b*^{2} – *ac*)*t*(*c*^{2} – *ab*)) = *a**t*^{2}(*a*^{2} – *bc*)^{2} – *t*^{2}(*b*^{2} – *ac*)(*c*^{2} – *ab*) = *a**t*^{2}[(*a*^{2} – *bc*)^{2} – (*b*^{2} – *ac*)(*c*^{2} – *ab*)] = *a*

Now it is just a matter of simplifying and solving for *t*.

*t*^{2}[*a*^{4} – 2*a*^{2}*bc* + *bc*^{2} – *bc*^{2} + *a**b*^{3} + *a**c*^{2} – *a*^{2}*bc*] = *a**t*^{2}[*a*^{4} – 3*a*^{2}*bc* + *a**b*^{3} + *a**c*^{3}] = *a*

Since *a* ≠ 0 we can divide both sides by *a* to get:

*t*^{2}(*a*^{3} + *b*^{3} + *c*^{3} – 3*a**bc*) = 1*t* = ±1/√(*a*^{3} + *b*^{3} + *c*^{3} – 3*a**bc*)

Thus we have the solution set:

(*x*, *y*, *z*)

= ±(*a*^{2} – *bc*, *b*^{2} – *ac*, *c*^{2} – *ab*)/√(*a*^{3} + *b*^{3} + *c*^{3} – 3*a**bc*)

To see how it works, consider a specific system:

*x*^{2} – *yz* = 2*y*^{2} – *xz* = 3*z*^{2} – *xy* = 5

We have *a* = 2, *b* = 3, *c* = 5, so we can evaluate:

*a*^{2} – *bc* = -11*b*^{2} – *ac* = -1*c*^{2} – *ab*) = 19

√(*a*^{3} + *b*^{3} + *c*^{3} – 3*a**bc*) = √70

Thus we have the two solutions:

*x* = -11/√70*y* = -1/√70*z* = 19/√70

*x* = 11/√70*y* = 1/√70*z* = -19/√70

And that’s the answer!

**References**

Toppr

https://www.toppr.com/ask/en-ca/question/solve-x2-yz-a2-y2-zx/

Math StackExchange

https://math.stackexchange.com/questions/3960053/solve-three-variable-system-x2-yz-1-y2-xz-2-z2-xy-3

YouTube community post

https://www.youtube.com/post/Ugkxy-uD3uApCj0LwiMcZYznSNQj34VAXQ7v

Cross product idea

https://math.stackexchange.com/questions/2129989/how-to-get-ratio-of-a-b-c-from-2-equations-in-a-b-c

### MY BOOKS

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

Book ratings are from January 2023.

(US and worldwide links)

https://mindyourdecisions.com/blog/my-books

**Mind Your Decisions** is a compilation of 5 books:

(1) The Joy of Game Theory: An Introduction to Strategic Thinking

(2) 40 Paradoxes in Logic, Probability, and Game Theory

(3) The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias

(4) The Best Mental Math Tricks

(5) Multiply Numbers By Drawing Lines

**The Joy of Game Theory** shows how you can use math to out-think your competition. (rated 4.3/5 stars on 290 reviews)

**40 Paradoxes in Logic, Probability, and Game Theory** contains thought-provoking and counter-intuitive results. (rated 4.2/5 stars on 54 reviews)

**The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias** is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 4.1/5 stars on 33 reviews)

**The Best Mental Math Tricks** teaches how you can look like a math genius by solving problems in your head (rated 4.3/5 stars on 116 reviews)

**Multiply Numbers By Drawing Lines** This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.4/5 stars on 37 reviews)

**Mind Your Puzzles** is a collection of the three “Math Puzzles” books, volumes 1, 2, and 3. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory.

**Math Puzzles Volume 1** features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 112 reviews.

**Math Puzzles Volume 2** is a sequel book with more great problems. (rated 4.2/5 stars on 33 reviews)

**Math Puzzles Volume 3** is the third in the series. (rated 4.2/5 stars on 29 reviews)

### KINDLE UNLIMITED

Teachers and students around the world often email me about the books. Since education can have such a huge impact, I try to make the ebooks available as widely as possible at as low a price as possible.

Currently you can read most of my ebooks through Amazon’s “Kindle Unlimited” program. Included in the subscription you will get access to millions of ebooks. You don’t need a Kindle device: you can install the Kindle app on any smartphone/tablet/computer/etc. I have compiled links to programs in some countries below. Please check your local Amazon website for availability and program terms.

US, list of my books (US)

UK, list of my books (UK)

Canada, book results (CA)

Germany, list of my books (DE)

France, list of my books (FR)

India, list of my books (IN)

Australia, book results (AU)

Italy, list of my books (IT)

Spain, list of my books (ES)

Japan, list of my books (JP)

Brazil, book results (BR)

Mexico, book results (MX)

### MERCHANDISE

Grab a mug, tshirt, and more at the official site for merchandise: **Mind Your Decisions at Teespring**.