Algebraic equations are normally divided in to two major types they are Linear equations and nonlinear equations. Linear equation has the power value as 1. But non linear equation’s variable has the power value of more than 1.Linear equation produce a straight line on the graph. But the non linear equation does not produce straight line.

Solutions to equations are the points that made the equation true that made the equation work correctly. Normally, non linear algebraic equation problems are frequently accurately solvable, and if not they generally can be thoroughly understand through qualitative and **numerical analysis**.

## Procedure for systems of non linear equations:

The numerical methods for estimate of real solutions of a system of non linear equations, i.e., finding the roots of a vector function. When compared with the one-dimensional case, finding roots in the multidimensional case is much more complex. For example, in one-dimensional case one can comparatively easily bracket the roots of a given function (i.e., conclude the intervals, in which at least one root of the function lies) but there are no methods for bracketing of roots of general functions in the multidimensional case! Usually we even do not know whether a root (solution of the system) exists and whether it is unique.

## Example for systems of non linear equations:

**Solve the following systems non linear equation:
**

**y = x ^{2}**

**y = 18 – x ^{2}**

Solution:

y = x^{2} ———–equation 1

y = 18 – x^{2} ———– equation 2

y = y

Plugging equation 1 in equation 2,

x^{2} = 18 – x^{2}

x^{2} + x2 = 18 – x^{2} + x^{2}

Each of these sub-equations is true, but the last one is usefully new and different:

Solve this for the x-values that make the equation true:

x^{2} = 18 – x^{2}

2x^{2} = 18

x^{2} = 9

x = –3, +3

Then the solutions to the unique system will occur when x = –3 and when x = +3.

Plug the x-values in to either of the two original equations. plug the x-values into the first equation, since it’s the simpler of the two:

x = –3:

y = x^{2}

y = (–3)2 = 9

x = +3:

y = x^{2}

y = (+3)2 = 9

**Then the solutions are (x, y) = (–3, 9) and (3, 9).**