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Solving an equation symbolically means that expressions can be used for representing the solutions. For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial: + = Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem). This particular equation, however, may be written
Once y is also eliminated from the third row, the result is a system of linear equations in triangular form, and so the first part of the algorithm is complete. From a computational point of view, it is faster to solve the variables in reverse order, a process known as back-substitution. One sees the solution is z = −1, y = 3, and x = 2. So ...
Use variable change z = x + m/x. So, z 2 = x 2 + ... This is a cubic equation in y. Solve for y using any method for solving such equations (e.g. conversion to a ...
This equation is an equation only of y'' and y', meaning it is reducible to the general form described above and is, therefore, separable. Since it is a second-order separable equation, collect all x variables on one side and all y' variables on the other to get: (′) (′) =.