# Systems of Nonlinear Equations

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 $(\pm 3,2),(\pm 2,-3)$ Return to Topics Page