Solve system x+y=5,xz+yu=7,xz^2+yu^2=11,xz^3+yu^3=19



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Posted on (Answer #1)

Solve the system:


From the first equation we get y=5-x. Substituting into the second equation we get:

xz+(5-x)u=7 ==> xz+5u-xu=7 ==> `u=(7-xz)/(5-x)`

We can now write the third and fourth equations in terms of x and z:

`xz^2+(5-x)((7-xz)/(5-x))^2=11` A

`xz^3+(5-x)((7-xz)/(5-x))^3=19` B

From A we can solve for x:





Substituting into B we get:




Now substitute for x; multiply through by `(5z^2-14z+11)^2` and add like terms to get:




(1) `z=1 => x=6/(5-14+11)=3` =>y=2 and u=2

(2) `z=2 => x=6/(20-28+11)=2` =>y=3 and u=1

(3) `z=7/5 => x=6/(5(7/5)^2-14(7/5)+11)=5` =>y=1; but `u=(7-xz)/(5-x)` so this value for z yields `u=0/0` so this is not a solution.


The solutions are x=2,y=3,u=1,z=2 and x=3,y=2,u=2,z=1



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