# Systems of Nonlinear Equations

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## 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: ${\displaystyle 3x-y=-2{\text{ and }}2x^{2}-y=0}$

Solution:

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

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

## Solving by elimination(example)

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

Solution:

We can subtract the second equation form the first to get ${\displaystyle y^{2}+y=6}$. 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 ${\displaystyle (\pm 3,2),(\pm 2,-3)}$

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