#Systems of equations
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Hint: For the first notice the sum of the three eqs is (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)/2=0, x=y=z is out, so reduces to x+y+z=0, x(x^2+y(x+y))=2, y(y^2+x(x+y))=-9.
Hint: The second is (x-y+1)^2=0,(x+y+1)^2=0, so reduces to a linear system
Hint: the third is (x+1)(y+1)=1, (x^3+1)(y^3+1)=7^2, (-2,-2) is spotted easily also t^3+1=(t+1)(t^2-t+1), so solve (x^2-x+1)(y^2-y+1)=7^2, (x+1)(y+1)=1
Here is a solution for the first system
idk if it is the solution, I just know that it is a solution bc I plugged it back in and it worked
here is the work part 1
part 2: plug y and z in terms of x into the first equation and solve for x
somone check my post please
Hint for (3) continued: ((x^2-x+1)(y^2-y+1)-7^2)-((x+1)(y+1)-1)((x-2)(y-2)-5)=3(x+y-4)(x+y+4), so (3) also has the two solutions of x+y=4, (x+1)(y+1)=1 in addition to (-2,-2).
What’s unclear?
proofs
You have all the ingredients
its a different problem
Where?