Solve the given equation system x(z+1)=y^(2)],[(z+1)^(2)=xy],[x^(2)=y(z+1)]

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Write System of Equations: Write down the system of equations. $$ \begin{cases} x(z+1) = y^2 \ (z+1)^2 = xy \ x^2 = y(z+1) \end{cases} $$Express y in terms: From the first equation, express y in terms of x and z. $$ y^2 = x(z+1) $$ $$ y = \sqrt{x(z+1)} $$ Note that we consider only the positive square root because if y were negative, squaring it would give a positive result, which would not match the original equation.Substitute in Second Equation: Substitute the expression for y from Step 2 into the second equation. $$ (z+1)^2 = x\sqrt{x(z+1)} $$Square to Eliminate Root: Square both sides of the equation from Step 3 to eliminate the square root. $$ ((z+1)^2)^2 = (x\sqrt{x(z+1)})^2 $$ $$ (z+1)^4 = x^2x(z+1) $$ $$ (z+1)^4 = x^3(z+1) $$Substitute y in Third Equation: From the third equation, we have $ x^2 = y(z+1) $. Substitute the expression for y from Step 2 into this equation. $$ x^2 = \sqrt{x(z+1)}(z+1) $$Square to Eliminate Root: Square both sides of the equation from Step 5 to eliminate the square root. $$ (x^2)^2 = (\sqrt{x(z+1)}(z+1))^2 $$ $$ x^4 = x(z+1)(z+1)^2 $$ $$ x^4 = x(z+1)^3 $$Equations Involving x and z: We now have two equations involving x and z: $$ (z+1)^4 = x^3(z+1) $$ $$ x^4 = x(z+1)^3 $$ We can use these equations to find a relationship between x and z.Divide Second Equation by x: Divide the second equation by x to simplify it, assuming x is not zero. $$ x^3 = (z+1)^3 $$Equate x and z+1: Since both $ x^3 $ and $ (z+1)^3 $ are equal, we can equate x and z+1. $$ x = z+1 $$Substitute x to find y: Substitute $ x = z+1 $ into the first equation to find y. $$ y^2 = (z+1)(z+1) $$ $$ y^2 = (z+1)^2 $$ $$ y = z+1 $$Substitute x to find z: Now we have $ x = z+1 $ and $ y = z+1 $. Substitute $ x = z+1 $ into the third equation to find z. $$ (z+1)^2 = (z+1)(z+1) $$ This is a tautology and does not give us new information about z. We need to find another way to determine z.Substitute x and y in Second Equation: Substitute $ x = z+1 $ and $ y = z+1 $ into the second equation to find z. $$ (z+1)^2 = (z+1)(z+1) $$ This is the same tautology as before. We have reached a point where all equations are satisfied for any z. Therefore, the system has infinitely many solutions, where $ x = y = z+1 $ for any real number z.

Solve the given equation system $x(z+1)=y^{2}, \ (z+1)^{2}=xy, \ x^{2}=y(z+1)$

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Solve the given equation system x(z+1)=y2, (z+1)2=xy, x2=y(z+1)x(z+1)=y^{2}, \ (z+1)^{2}=xy, \ x^{2}=y(z+1) x(z+1)=y2, (z+1)2=xy, x2=y(z+1)

Solve the given equation system $x(z+1)=y^{2}, \ (z+1)^{2}=xy, \ x^{2}=y(z+1)$