How find the value of the $x+y$

Question

Question:

let $x,y\in \Bbb R $, and such $$\begin{cases} 3x^3+4y^3=7\\ 4x^4+3y^4=16 \end{cases}$$

Find the $x+y$

This problem is from china some BBS

My idea: since $$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$ $$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$

then I can't Continue

Answer 1

Assuming the question is typed correctly as shown, there are two unique real solutions. Let $$\begin{align*} u(z) &= -983749-111132z^3+786432z^4+71442z^6-196608z^8-20412z^9+18571z^{12}, \\ v(z) &= -178112-351232z^3+186624z^4+301056z^6-34992z^8-114688z^9+18571z^{12}. \end{align*}$$ These polynomials have exactly two distinct real roots; let $r(u,+)$, $r(u,-)$ be the positive and negative real roots of $u$, and $r(v,+)$, $r(v,-)$ be the positive and negative real roots of $v$, respectively. Then $$(x,y) \in \{(r(u,-),r(v,+)), (r(u,+),r(v,-))\}$$ are the desired solutions. The sum $x+y$ can then be expressed by the solution to a third polynomial $$f(z) = 819447-537600z-8998752z^3+3291428z^3+22132992z^4-17875200z^5+3163146z^6+1042512z^8-437500z^9+18571z^{12},$$ for which there are again two real roots, both positive. All of these polynomials are irreducible. So I highly doubt that this is a problem that can be reasonably solved by hand.

Answer 2

Here comes a drawing from a credible (Delphi Pascal) source (-:

Viewport:

  xmin := -4; xmax := +4;
  ymin := -4; ymax := +4;

Color assignment: $$ \begin{cases} \color{red}{3x^3+4y^3=7}\\ \color{green}{4x^4+3y^4=16} \end{cases} $$ Hope this helps. (Now thinking about the rest ..)

EDIT. A parameter representation for the curve $\;3x^3+4y^3=7\;$ is: $$ x(t) = -\sqrt[3]{\frac{t}{3}} \qquad ; \qquad y(t) = \sqrt[3]{\frac{t+7}{4}} $$ This turns the problem into seeking zeroes of a one-dimensional function: $$ f(t) = 4 x^4(t) + 3 y^4(t) - 16 $$ Standard numerical methods can be employed for this purpose. Newton-Raphson and Regula Falsi have been successful here. Our (double precision) results are: $$ t = 3.56874527617414 \quad \Longrightarrow \quad (x,y) = ( -1.05957433921527 , +1.38246606572819 ) \\ t = -8.23503156460535 \quad \Longrightarrow \quad (x,y) = ( +1.40017183631184 , -0.675884813969605 ) $$ Giving, respectively: $$ x+y = 0.322891726512912 \\ x+y = 0.724287022342236 $$ ALGORITHM (Delphi Pascal) :

program RF;

type
  funktie = function(t : double) : double;

function regula_falsi(d1,d0,eps : double; F : funktie) : double;
{
  Regula Falsi
}
var
  OK : boolean;
  d2 : double;
begin
  d2 := d0;
  while abs(F(d2)) > eps do
  begin
    d2 := d1 - F(d1)*(d0 - d1)/(F(d0) - F(d1));
    if F(d2) = 0 then Break;
    OK := (F(d2)F(d1) < 0);
    if not OK then
    begin
      d1 := d2;
    end else begin
      d0 := d1; d1 := d2;
    end;
    Write(d2,' : ',F(d2),' ; Press Enter'); Readln;
  end;
  regula_falsi := d2;
end;

function power(x,r : double) : double;
var
  M : double;
begin
  M := 0;
  if x > 0 then
    M := exp(rln(abs(x)));
  if x < 0 then
    M := -exp(r*ln(abs(x)));
  power := M;
end;

function original(t : double) : double;
{
  Function itself
}
var
  x,y,f : double;
begin
  x := -power(t/3,1/3);
  y :=  power((t+7)/4,1/3);
  f := 4*sqr(sqr(x))+3*sqr(sqr(y))-16;
  original := f
end;

procedure Calculate(x1,x2 : double);
const
  eps : double = 1.E-14;
var
  t,x,y,t1,t2 : double;
begin
  t1 := 3*sqr(x1)*x1;
  t2 := 3*sqr(x2)*x2;
  t := Regula_Falsi(t1,t2,eps,original);
  Writeln(t);
  x := -power(t/3,1/3);
  y :=  power(t/4+7/4,1/3);
  Writeln(x,' +',y,' =',x+y); 
end;

begin
  Calculate( 1.0, 1.5);
  Writeln;
  Calculate(-1.5,-0.5);
end.

Note that the only place that is truly machine dependent is the stopping criterion with 'eps'. As for the rest, the algorithm is quite general. Also take note of the rather crude starting values $(1.0,1.5)$ and $(-1.5,-0.5)$ for $x$ at the end of the code, emphasizing the robustness of this algorithm.

Answer 3

Another idea for you to think about. We have:

$$\begin{cases} 3x^3+4y^3=7 \\ 4x^4+3y^4=16 \end{cases}$$

Break it as:

$$\begin{cases} 3x^3+3y^3 + y^3=7 \\ x^4 + 3x^4+3y^4=16 \end{cases}$$

and factor:

$$\begin{cases} 3(x+y)(x^2 - xy + y^2)=7 - y^3\\ 3(x+y)(x^3 - x^2y + xy^2 - y^3)=16 - x^4 \end{cases}$$

And so: $$x+y = \frac{1}{3}\frac{7 - y^3}{x^2 - xy + y^2} = \frac{1}{3}\frac{(2-x)(8+4x+2x^2+x^3)}{x^3 - x^2y + xy^2 - y^3}$$

Answer 4

Defining $z:= x+y$, the equations become $$\begin{cases} 3x^3+4(z-x)^3=\phantom{1}7\\ 4x^4+3(z-x)^4=16 \end{cases}$$

Using the method of resultants (which could conceivably be done "by hand", but I used Mathematica because life's too short) to eliminate $x$, we confirm @heropup's result:

$$\begin{align} 0 &= 819447 - 537600 z - 8998752 z^2 + 3291428 z^3 + 22132992 z^4 \\ &- 17875200 z^5 + 3163146 z^6 + 1042512 z^8 - 437500 z^9 + 18571 z^{12} \quad (\star) \end{align}$$

Solving an irreducible $12$th-degree polynomial equation is tricky at best. Again using Mathematica, the two real roots of $(\star)$ are found to be $$z=0.322892\dots \quad\text{and}\quad z=0.724287\dots$$

If we abstract the equations like this ... $$\begin{cases} a_1 x^3 + b_1 (z-x)^3=c_1\\ a_2 x^4 + b_2 (z-x)^4=c_2 \end{cases}$$ ... then the corresponding resultant is ... $$\begin{align}0 &= c_1^4\;(a_2 + b_2)^3 - c_2^3\;(a_1 - b_1)^4 \\ &- 12 z \; (a_1 - b_1)^2 \; (a_1 b_2 + a_2 b_1)\; c_1 c_2^2 \\ &+ \cdots \\ &- 24 z^5\;(a_1 b_2 + a_2 b_1) \; (a_1 a_2 b_1 + 2 a_2 b_1^2 + 2 a_1^2 b_2 + a_1 b_1 b_2) \; c_1 c_2 \\ &+ \cdots \\ &- 4 z^9 \; ( a_1 b_2 + a_2 b_1)^3 \;c_1 \\ &+ z^{12}\; ( a_1^4 b_2^3 + a_2^3 b_1^4 ) \end{align}$$

Observations:

• A number of coefficients vanish under the assumption $a_1 b_2 + a_2 b_1 = 0$, although this doesn't really help much, because the polynomial remains irreducible still and degree $12$.

• We can reduce the degree to $9$ with the relation $a_1^4 b_2^3 + a_2^3 b_1^4 = 0$, but that's also not particularly helpful.

• We can eliminate the constant term if $c_1 = |a_1 - b_1| c^3$ and $c_2 = (a_2 + b_2) c^4$ for some $c$. In the original problem, with $a_1 = b_2 = 3$ and $b_1 = a_2 = 4$, this gives $c_1 = c^3$ and $c_2 = 7 c^4$. If $b_2$ had a sign-change to $-3$, then $c_1 = c^3$ and $c_2 = c^4$, which for $c=2$, give $8$ and $16$ ... almost the $7$ and $16$ of the original problem. Even so, polynomial $(\star)$ only reduces to $11$th degree.

If there's something remarkable about the solutions to this problem, I'm not seeing it.