Systems of Nonlinear Equations

Comparison to linear systems

Systems of nonlinear equations are solved using the same methods we used to solve linear systems, elimination and substitution.

Solving by substitution (example)

Solve the following system of equations: $3x-y=-2{\text{ and }}2x^{2}-y=0$

Solution:

Using the first equation, we see that y = 3x + 2. Substituting 3x + 2 for y in the second equation we see that $2x^{2}-(3x+2)=0$ Now we can solve for x, by using the quadratic formula, or factoring. We find that $(2x+1)(x-2)=0$ or that $x={\frac {-1}{2}},2$

Solving back for y, we find that the two points on both curves are $({\frac {-1}{2}},{\frac {1}{2}}),{\text{ and }}(2,8)$

Solving by elimination(example)

Solve the following system by elimination: $x^{2}+y^{2}=13{\text{ and }}x^{2}-y=7$

Solution:

We can subtract the second equation form the first to get $y^{2}+y=6$. We can solve this equation for y to find that y = 2 or -3. For each value of y, we have 2 values for x. So we have four points of intersection $(\pm 3,2),(\pm 2,-3)$

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