### Graphs of Nonlinear Systems

A nonlinear system of equations includes at least one equation that is not linear. The nonlinear equation(s) may have at least one variable raised to a power greater than 1 or a product of variables. Graphing the equations in the system gives a visual representation of how many solutions the system has. The possible number of solutions in a system depends on the types of equations in the system.

For example, the graph of a system with one linear equation and one quadratic equation includes a line and a parabola. If they do not intersect, there are no solutions. If they intersect in one point, there is one solution. If they intersect in two points, there are two solutions. In some cases, exact solutions can be identified from the graph. These solutions should be verified by substituting them back into the original equations. In some systems, a graph may only give an approximate solution.Graph each equation.

The first equation is:The solutions are $(-4,0)$ and $(0,4)$.

To check that the points of intersection are solutions, substitute the values for $x$ and $y$ back into both equations and simplify. If they both result in true equations, then the ordered pairs are solutions.

Check $(-4,0)$:### Solving Nonlinear Systems Algebraically

Solve one of the equations for one of the variables.

If possible, choose a variable that is not squared or raised to a higher power.

Solve for $y$.Substitute the expression from Step 1 for the variable in the other equation.

Substitute $-\frac{5x}{2}$ for $y$ in the second equation:The solutions are $(2, -5)$ and $(-2,5)$.

To check the solutions, substitute the pairs of values for $x$ and $y$ back into both equations and simplify. If they both result in true equations, then the solutions are correct.

Check $(2,-5)$:Substitute the values from Step 3 into either of the original equations and then solve for the other variable.

Substitute –1.5 and 3 for $x$.The solutions are $(3,\pm 4)$ and $(-1.5, \pm \sqrt{22.75})$.

The solutions can be checked by substituting the values for $x$ and $y$ back into the original equations. If a solution makes both equations true, then the solution is correct.