Pair of equations both of degree 2?

[JSimmonds49]

Hi,

I have been given these two equations and have been unable to solve them as of yet.

x^2+(y(sqrt(xy)))= 2013
y^2+(x(sqrt(xy)))= 671

I have found out so far that 3 * 671 is 2013 and the equation could potentially be solved using the fact sqrt(xy) is in both equations, I need to express the value of y^2 as a fraction a/b, where a and b are positive coprime numbers. Also, the value of a+b does not exceed 999

Can anyone help?

 

[JSimmonds49]

Formatting...

 

[MarkFL]

Try the substitution \(\displaystyle x=9y\), and you find the two equations become equivalent.

 

[Deleted member 4993]

View attachment 3148

Assume

x = my

then

m² y² + y²m^1/2 = 2013 → y² (m² + m^1/2) = 3*671 ........................(1) edited

y² + my*ym^1/2 = 671 → y² (1 + m^3/2) = 671 ............................(2)

m² + m^1/2 = 3 * (1 + m^3/2 )

let

n = m^1/2

n⁴ + n = 3 * (1 + n³ )

n⁴ - 3n³ + n - 3 = 0

n³(n-3) + 1(n-3) = 0 → (n - 3)(n³ + 1) = 0 → n = 3 → m = 9 → x = 9y...............edited (Thanks Mark)

since the equations are not independent, we can have the equation above only.

 

[MarkFL]

Assume

x = my

then

m² y² + y²m^1/2 = 2013 → y² (m² + m^1/2) = 3*671 ........................(1)

y² + my*ym^1/2 = 671 → y² (1 + m^3/2) = 671 ............................(2)

m² + m^1/2 = 3(1 + m^3/2)

let

n = m^1/2

n⁴ + n = 3(1 + n³ )

n⁴ - 3n³ + n - 3 = 0

n³(n - 3) + 1(n -3) = 0 → (n - 3)(n³ + 1) = 0 → n = 3 → m = 9 → x = 9y

since the equations are not independent, we can have the equation above only.

You switched the two constants in the original equations.

 

[JSimmonds49]

Almost there...

Thanks for the help guys, I need a numeric value for y^2 now as fraction a/b, so a+b is an integer between 0 and 999.

 

[MarkFL]

Thanks for the help guys, I need a numeric value for y^2 now as fraction a/b, so a+b is an integer between 0 and 999.

Use the substitution \(\displaystyle x=9y\) in either original equation, and you will get exactly what you need.

 

[JSimmonds49]

I've tried substituting x=9y into the second equation and got:

y^2+9y(sqrt(9y^2))

28y^2=671

This doesn't give an integer solution, where am I going wrong?

 

[Deleted member 4993]

I've tried substituting x=9y into the second equation and got:

y^2+9y(sqrt(9y^2))

28y^2=671

This doesn't give an integer solution, where am I going wrong?

Why do you think y² needs to be an integer!!

Re-read your question again.

 

[MarkFL]

You never mentioned that the equations were Diophantine. But, what you get is:

\(\displaystyle y^2=\dfrac{671}{28}\)

and:

\(\displaystyle 671+28\le999\)

This is the type of solution you expressed needing.

 

[JSimmonds49]

Got it!

Got the answer, thanks guys!

28y^2=671
y^2=671/28
671+28=699

 

[Dale10101]

A couple of questions



I hope it is okay to ask second party questions.
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