System of Non-linear Equations

[mojacko]

Find the product of two numbers such that their sum multiplied by the sum of their squares is 65, and their difference multiplied by the difference of their square is 5.

help me to answer this problem,
i tried to solve this using substitution and elimination, but im stuck,

Let x = the first number
y = the second number

(x+y)(x^2+y^2) = 65
(x-y)(x^2-y^2) = 5

then i expand both equation: x^3+xy^2+yx^2+y^3 = 65
x^3-xy^2-yx^2+y^3= 5

then i add eq. 1 and eq. 2: then i got 2x^3+y^3=70 ..
i still have two unknown, i cant eliminate one of the variable .. help me please ..

thanks ..

 

[stapel]

Find the product of two numbers such that their sum multiplied by the sum of their squares is 65, and their difference multiplied by the difference of their square is 5.

Yeesh! This is nasty!

i tried to solve this using substitution and elimination, but im stuck,

Let x = the first number
y = the second number

(x+y)(x^2+y^2) = 65
(x-y)(x^2-y^2) = 5

Thank you for showing your work so nicely!

At the last line of your work (shown above), I would try factorization:

. . . . .(x - y)(x - y)(x + y) = 5

Now divide:

. . . . .[(x + y)(x^2 + y^2)] / [(x + y)(x - y)^2] = 65/5

. . . . .(x^2 + y^2) / (x - y)^2 = 13

Multiply out:

. . . . .(x^2 + y^2) / (x^2 - 2xy + y^2) = 13

. . . . .x^2 + y^2 = 13x^2 - 26xy + 13y^2

. . . . .0 = 12x^2 - 26xy + 12y^2

. . . . .0 = 6x^2 - 13xy + 6y^2

Can you factor this? And then see where that leads?

 

[JeffM]

Find the product of two numbers such that their sum multiplied by the sum of their squares is 65, and their difference multiplied by the difference of their square is 5.

help me to answer this problem,
i tried to solve this using substitution and elimination, but im stuck,

Let x = the first number
y = the second number

(x+y)(x^2+y^2) = 65
(x-y)(x^2-y^2) = 5

then i expand both equation: x^3+xy^2+yx^2+y^3 = 65
x^3-xy^2-yx^2+y^3= 5

then i add eq. 1 and eq. 2: then i got 2x^3+y^3=70 ..
i still have two unknown, i cant eliminate one of the variable .. help me please ..

thanks ..

Did you try graphing both equations?

Look in the first quadrant. Do you see any points of intersection? I do.

Whenever you have some problem that is hard to deal with analytically, think about graphing. It may give you the answer (or at least an approximate answer). It may give you a clue on how to proceed analytically.

 

[mojacko]

Yeesh! This is nasty!


Thank you for showing your work so nicely!

At the last line of your work (shown above), I would try factorization:

. . . . .(x - y)(x - y)(x + y) = 5

Now divide:

. . . . .[(x + y)(x^2 + y^2)] / [(x + y)(x - y)^2] = 65/5

. . . . .(x^2 + y^2) / (x - y)^2 = 13

Multiply out:

. . . . .(x^2 + y^2) / (x^2 - 2xy + y^2) = 13

. . . . .x^2 + y^2 = 13x^2 - 26xy + 13y^2

. . . . .0 = 12x^2 - 26xy + 12y^2

. . . . .0 = 6x^2 - 13xy + 6y^2

Can you factor this? And then see where that leads?

6x^2 -13xy + 6y^2 = 0
6x^2 - 9xy - 4xy + 6y^2 = 0
3x(2x - 3y) - 2y (2x - 3y) = 0
(3x - 2y)(2x - 3y) = 0

thank you very much for your help!

 

[mojacko]

That sould be 2x^3 + 2y^3 = 70, so x^3 + y^3 = 35
Gives you an "eyesight" solution of (x,y) = (3,2) or (2,3)

Subtracting the equations (instead of adding) gives: (y)x^2 + (y^2)x - 30 = 0, a quadratic.

Play with that...

actually that's a typo error, but anyways thanks for your suggestions, i also tried that and i got it now thank you

 

[Dale10101]

Plot

Yeesh! This is nasty!


Thank you for showing your work so nicely!

At the last line of your work (shown above), I would try factorization:

. . . . .(x - y)(x - y)(x + y) = 5

Now divide:

. . . . .[(x + y)(x^2 + y^2)] / [(x + y)(x - y)^2] = 65/5

. . . . .(x^2 + y^2) / (x - y)^2 = 13

Multiply out:

. . . . .(x^2 + y^2) / (x^2 - 2xy + y^2) = 13

. . . . .x^2 + y^2 = 13x^2 - 26xy + 13y^2

. . . . .0 = 12x^2 - 26xy + 12y^2

. . . . .0 = 6x^2 - 13xy + 6y^2

Can you factor this? And then see where that leads?

Interesting what this last equation does and does not give you.

(click image for larger version.)




It is enlightening to realize that given the original two equations one looks for a solution by transposition and direct substitution but that doesn't work, a solution by factoring and simpliying each equation by itself but that doesn't work. One is left with combing the two equations ... and what are the choices (?), interestingly, the two equations, just like numbers, can be added, subtracted, divided or multiplied by one another to get a single equation ... but so what, what does one hope for ... if you are very lucky to cancel out x or y ... but mostly likely you are left with another two variable equation just as you started with. It is the last step of the problem that shows what the goal was in combing the two equations (as the graph discloses) ... not a new equation that provides a solution set itself but a means to eliminate a variable from one of the two original equation. Very interesting, one of the two solutions is (2,3) and checking this solution in the first equation shows a symmetry which discloses what the second solution must be. Most educational, this problem.