When a single variable is specified and a particular root of an equation has multiplicity greater than one, Solve gives several copies of the corresponding solution. Solve [ expr , vars ]
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just use the Solve function. In your case it becomes : Solve[x + y == 2 x, y] that gives: {{y -> x}}
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Solve[x == y && x == -y, x, {y}] which is equivalent to. Solve[Exists[y, x == y && x == -y], x] {{x -> 0}} Back to your case. Conclusion: either use reduce or add more symbols to the variable list. I can
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Solve equation
Try to find the answer to the equation.
Solving Systems Of Linear Equations One Step At A Time Wolfram Alpha Blog. Equation Solving In The Wolfram Age Mathematica You. Solve equations in mathematica