help please w/ Simultaneous NON-linear equations: 4xy = -15, 4x^3+4y^3 =49

[Barry247]

Hi all I am new to this forum!
Could anybody help me with this simultaneous equation problem, I have tried in vain to solve this for many hours but to no avail

4xy = -15

4x^3+4y^3 =49

Steps so far

y= -15/4x

4x^3+4(-15/4x)^3= 49

4x^3 +4(-3375/64x^3) = 49

4x^3 -13500/64x^3) =49

64x^3(4x^3) -13500 =49x^3

256x^6 -13500 = 49x^3

256x^6 - 49x^3 -13500 = 0


This is as far as I can get


I have tried every way I know but it does not seem to solve, I must be missing something I have solved 17 out of 18 equations from my a level maths text book but this one seems to have beaten me
I would appreciate any help

Many thanks in advance

Barry

 

[tkhunny]

Hi all I am new to this forum!
Could anybody help me with this simultaneous equation problem, I have tried in vain to solve this for many hours but to no avail

4xy = -15

4x^3+4y^3 =49


I have tried every way I know but it does not seem to solve, I must be missing something I have solved 17 out of 18 equations from my a level maths text book but this one seems to have beaten me
I would appreciate any help

Many thanks in advance

Barry

What's preventing you from solving the first for x or for y? Did you do that? You should tell us if you did.

Are you SURE there IS a solution? With xy = -15/4, you'll find things possible only in Quadrants II and IV (Assuming you're interested in only Real solutions). Why?

 

[Harry_the_cat]

Hi all I am new to this forum!
Could anybody help me with this simultaneous equation problem, I have tried in vain to solve this for many hours but to no avail

4xy = -15

4x^3+4y^3 =49

Steps so far

y= -15/4x

4x^3+4(-15/4x)^3= 49

4x^3 +4(-3375/64x^3) = 49

4x^3 -13500/64x^3) =49

64x^3(4x^3) -13500 =49x^3 … this should be 49 * 64 x^3 on RHS

256x^6 -13500 = 49x^3

256x^6 - 49x^3 -13500 = 0 … let a = x^3, this is then a quadratic in "a", solve for a, then solve for x


This is as far as I can get


I have tried every way I know but it does not seem to solve, I must be missing something I have solved 17 out of 18 equations from my a level maths text book but this one seems to have beaten me
I would appreciate any help

Many thanks in advance

Barry

see red

 

[Dr.Peterson]

Hi all I am new to this forum!
Could anybody help me with this simultaneous equation problem, I have tried in vain to solve this for many hours but to no avail

4xy = -15

4x^3+4y^3 =49

Steps so far

y= -15/4x

4x^3+4(-15/4x)^3= 49

4x^3 +4(-3375/64x^3) = 49

4x^3 -13500/64x^3) =49

64x^3(4x^3) -13500 =49x^3

256x^6 -13500 = 49x^3

256x^6 - 49x^3 -13500 = 0


This is as far as I can get


I have tried every way I know but it does not seem to solve, I must be missing something I have solved 17 out of 18 equations from my a level maths text book but this one seems to have beaten me
I would appreciate any help

Many thanks in advance

Barry

Make the substitution u = x^3.

But check your work, too; I did the same things, but had very different coefficients. I think you failed to multiply every term by the entire denominator.

 

[Barry247]

Tried this so far

Make the substitution u = x^3.

But check your work, too; I did the same things, but had very different coefficients. I think you failed to multiply every term by the entire denominator.

many thanks for you quick response

Is this correct way to go

y = -15/4x

4x^3 + 4(-15/4x)^3 = 49

4x^3 + 4(-3375/64x^3) =49

4x^3 - 13500/64x^3 =49

(64x^3)(4x^3) -13500 = 49 (64x^3)


a = x^3

(64a)(4a) -13500 =49(64a)


256a^2 -3136a -13500 = 0


this does not seem to cancel down to smaller numbers


Is this correct so far

Thanks in advanced

 

[Steven G]

Hi all I am new to this forum!
Could anybody help me with this simultaneous equation problem, I have tried in vain to solve this for many hours but to no avail

4xy = -15

4x^3+4y^3 =49

Steps so far

y= -15/4x

4x^3+4(-15/4x)^3= 49

4x^3 +4(-3375/64x^3) = 49

4x^3 -13500/64x^3) =49

64x^3(4x^3) -13500 =49x^3

256x^6 -13500 = 49x^3

256x^6 - 49x^3 -13500 = 0


This is as far as I can get


I have tried every way I know but it does not seem to solve, I must be missing something I have solved 17 out of 18 equations from my a level maths text book but this one seems to have beaten me
I would appreciate any help

Many thanks in advance

Barry

As pointed out, you did not multiply both sides by the same factor. Whenever the powers are a multiple of 3 (or 4 or 129) you should replace x^3 with u (as already mentioned by Dr P) (x^4 with u or x^129 with u).

Do this and come back showing your work.

 

[Deleted member 4993]

many thanks for you quick response

Is this correct way to go

y = -15/4x

4x^3 + 4(-15/4x)^3 = 49

4x^3 + 4(-3375/64x^3) =49

4x^3 - 13500/64x^3 =49

(64x^3)(4x^3) -13500 = 49 (64x^3)


a = x^3

(64a)(4a) -13500 =49(64a)


256a^2 -3136a -13500 = 0


this does not seem to cancel down to smaller numbers


Is this correct so far

Thanks in advanced

That is a quadratic equation with a quadratic function.

What are the roots of this quadratic function?

 

[Dr.Peterson]

many thanks for you quick response

Is this correct way to go

y = -15/4x

4x^3 + 4(-15/4x)^3 = 49

4x^3 + 4(-3375/64x^3) =49

4x^3 - 13500/64x^3 =49

(64x^3)(4x^3) -13500 = 49 (64x^3)


a = x^3

(64a)(4a) -13500 =49(64a)


256a^2 -3136a -13500 = 0


this does not seem to cancel down to smaller numbers


Is this correct so far

Thanks in advanced

Your equation is correct now. It will turn out that it could in principle be factored, but I'd definitely use the quadratic formula. (You can divide everything by 4 to simplify it, but by nothing more than that.)

 

[Barry247]

Solution found many thanks

Your equation is correct now. It will turn out that it could in principle be factored, but I'd definitely use the quadratic formula. (You can divide everything by 4 to simplify it, but by nothing more than that.)

I have now solved this equation using the quadratic formula for values of x, many thanks for you help

Barry