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

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Solve the system:

x+y=5
xz+yu=7
`xz^2+yu^2=11`
`xz^3+yu^3=19`

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:

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

`=>49-14xz+x^2z^2=55-11x-5xz^2+x^2z^2`

`=>5xz^2-14xz+11x-6=0`

`=>x=6/(5z^2-14z+11)`

Substituting into B we get:

`(7-xz)^3/(5-x)^2=19-xz^3`

`=>(7-xz)^3=(19-xz^3)(5-x)^2`

`=>-10x^2z^3+25xz^3+21x^2z^2-147xz-19x^2+190x-132=0`

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

`750z^5-5400z^4+15360z^3-21588z^2+14994z-4116=0`

`=>6(z-2)(z-1)(5z-7)^3=0`

`=>z=2,1,7/5`

(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.

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

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