How to solve a hard nonlinear system of equation

Here's another system of nonlinear equations! I got this question from the Math League math contest. We are given x+xy+y=11 and x^2y+xy^2=30 and will solve for both x and y. Do not immediately solve this system of equations by elimination or substitution! There's a nice trick! : )

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#math #algebra #mathbasics

[MichaelPennMath]

Dominating the channel diversification game!

[phamnguyenductin]

We Vietnamese students call this a basic "type 1 symmetric system of equations", which is when the equations remain the same if the two variables are interchanged.
The standard way to solve these systems is to substitute S = x + y and P = xy, solve for S and P, and use Vieta's formula to solve for x and y.

[cynthia]

GOODNESS! So, I went the REALLY long way around by substituting from the start. I ended up with a fourth degree polynomial that I had to divide synthetically. PHEW! I had fun doing it, but you always amaze me with your methods. Part of me is mad that you made it so easy. haha I love it

[plislegalineu]

When you were doing the systems of equations with a and b, and x any y (not the Q itself) you could multiply the multiplication (ab or xy) by 2 and then add or subtract the equations to get a perfect square

[GirishManjunathMusic]

Before watching the video:

Given:
x + y + xy = 11
x²y + xy² = 30

xy(x + y) = 30
xy = 30/(x + y)

x + y + xy = 11
x + y + 30/(x + y) = 11
(x + y)² + 30 = 11(x + y)
(x + y)² –11(x + y) + 30 = 0
(x + y) = (11 ± √(121 – 120))/2
(x + y) = (11 ± 1)/2
(x + y) = 5 or 6.
correspondingly, from xy = 30/(x + y),
xy = 6 or 5.
((x + y), (xy)) = (5, 6) or (6, 5)

x + y = 6
xy = 5
x(6 – x) = 5
x² – 6x + 5 = 0
x = (6 ± √(36 – 20))/2
x = (6 ± 4)/2
x = 1 or 5
Correspondingly, y = 5 or 1.

x + y = 5
xy = 6
x(5 – x) = 6
x² –5x + 6 = 0
x = (5 ± √(25 – 24))/2
x = (5 ± 1)/2
x = 2 or 3
Correspondingly, y = 3 or 2

(x, y) = (1, 5), (5, 1), (2, 3), or (3, 2)

[brrrrrrruh]

After the substitution a=x+y, b=xy you get a symmetric system which can have at most 2 roots (rearranging the first eq for a and substituting will give you a quadratic), thus you can guess just 1 solution. It's clearly 5, 6 and 6, 5. Now you have 2 systems in x and y of the same sort and for a, b= 5, 6 you guess x, y=1, 5, or 5, 1, for a, b=6, 5 you guess 2, 3 and 3, 2. So actually this can be solved in 30 seconds without even pen and paper!

[mesi]

you are amazing Mr I appreciate for your clarification.

[jarkola]

There's an easier way that I used, let (x+y) and xy be roots of quadratic equation z^2 - 11z + 30 = 0
We get that z = 6 or 5
Make cases for x+y and xy
Thus, a few more quadratic equations can be written
Solving them we get (x, y) € {(1, 5), (5, 1), (3, 2), (2, 3) }
Done. Easy!

[valvaraad]

Modify the second equation to be xy(x+y), then change xy=a and x+y=b.
Sorry for bad English, I just forgot the lexicon

[pNsB]

I was actually able to solve this pretty quickly:
Let s=x+y, and p=xy
s+p=11
sp=30
So s or p is equal to 5, and the other is 6. If p=5 and s=6, then x and y are 1 and 5. If s=5 and p=6, then x and y are 2 and 3.
Therefore the solutions are (1, 5), (2, 3), (3, 2) and (5, 1).

[not_vinkami]

I'm feeling good because all I needed to solve the question is the question itself, my brain, and a single finger that drew random stuff in the air

[MrRenanwill]

Multiplying the first line by x*y, we have:
x^2 y + x^2 y^2+xy^2= 11xy
Hence, by the second line, the first line becomes:
30 + x^2 y^2 = 11 xy.
By the quadratic formula applyerd for z=xy
xy = (11 +/- sqrt( 11^2-4*30 ))/2

Now we have x in terms of y. I would try the first equation.

[jakistam]

Before watching: I claim that the following pairs: (5, 1), (1, 5), (2, 3), (3, 2) are all of the solutions.

I was correct. My method basically boiled down to the same as his, though I arrived there in a more roundabout way and had to go through more calculations.

[devondevon]

Ignore the second equation and focus only on the first x+xy+ y =11
since (x+1)(y+1)= x+xy+y+1; hence add 1 to both sides of the equation
x+xy+y+1 = 12
(x+1)(y+1) = 3*4 or 6*2 or 12*1
for 3*4
x+1 =3, hence x=2 y+1= 4 hence y=3 (2, 3) answer
for 6*2
x+1 = 6 ; hence x =5 y+1 = 2 ; hence y =1 (5, 1) answer

for 12*1
x+1 = 12 hence x=11 y+1 = 1 ;hence y =0 (11, 0) cant use this because of the 0
which would turn the second equation in to 0

hence the answer is 2, 3 and 5, 1

[NeedSpeeeeed]

There is a formula for this kind of systems
Solve x *2 -Sx+P=0
S : sum
P: product
Do it twice

[mmoglsp]

I rewrote it as
y=-2x+11
x^2(-2x+11)+x(-2x+11)=30
I then found three solutions, and assumed x=5, the only rational solution, to be the correct one.
f1:5+(5)(1)+1=11
f2:(25)(1)+(5)(1)=30
Therefore, (5, 1) satisfies the nonlinear system of equations.

[sonaruo]

second equation factor out the xy so you get xy(x+y)=30=2*3*5
so 3 numbers multiply gives 30 which can be broken down to 3 numbers and here you have it
2, 3 or 3.2

[Professor_Sargeant_JAMS]

Squaring both sides of Equation 1, x^2 y^2 + 2 x^2 y + x^2 + 2 x y^2 + 2 x y + y^2 = 121. Multiplying both sides of Equation 2, 2x^y + 2xy^2 = 60. Substituting, x^2 y^2 + x^2 + 2 x y + y^2 + 60 = 121. So, (xy)^2 + (x + y)^2 = 61. 61 is a prime number that is equivalent to 1 mod 4. Fermat tells us there can only be a unique expression if 61 as a sum of positive numbers being squared< up to arrangement. 61 = 25 + 36 = 5^2 + 6^2. The first part is a mess, but once we apply the information from Fermat, we are at xy = 5 and x + y = 6 or xy = 6 and x + y = 5. Sorry for replying 1 to 2 years late. Next question is how many in this family of questions could be dealt with this way.

[tamirerez]

I haven't decided yet what you are more talented at.
Solving difficult math problems, or changing colors while writing.😅👍❤️

[holyshit]

Factor LHS of second equation to get
xy(x+y)=30
From first equation calculate xy
11- (x+y)=xy
(11-(x+y))(x+y)=30
Let u=x+y
(11-u)u=30
-u^2+11u=30
u^2-11u+30=0
(u-5)(u-6)=0
We have two systems of equations
x+y=5
xy = 6
x+y=6
xy=5
These both system of equations are Vieta formulas for quadratics
t^2-5t+6=0 // For first system of equations
t^2-6t+5=0 // For the second system of equation
(x, y) = {(2, 3), (3, 2), (1, 5), (5, 1)}