 # How to Solve Systems of Nonlinear Equations

When solving systems of nonlinear equations, you use the same approach as for systems of equations. The substitution method is the one we’ll use here.

Remember that in these cases, we can get more than one solution.

Rule

### Systemsofnonlinearequations

1.
You’ll find $x$ or $y$ by using the simplest equation.
2.
Put one expression into the other expression. Here you can get up to two solutions.
3.
These two equations have to be put back into the first equation you used one-by-one. First, you put in one of the solutions and solve for the corresponding variable. Then, you put in the other solution and solve for this solution’s corresponding variable.
4.
You can have more than one solution: $\left({x}_{1},{y}_{1}\right)$, $\left({x}_{2},{y}_{2}\right)$ and so on.

Example 1

Solve the system of equations $\begin{array}{lll}\hfill y+2x& =-1\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill {x}^{2}+3x& =5-y\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\phantom{\rule{0.33em}{0ex}}\end{array}$

1.
Choose Equation (1) and solve for $y$: $\begin{array}{llll}\hfill y+2x& =-1,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =-1-2x.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
2.
Put this solution into Equation (2) and make it into an expression consisting only of $x$: $\begin{array}{llll}\hfill \phantom{\rule{-20.07507pt}{0ex}}{x}^{2}+3x& =5-y\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+3x& =5-\left(-1-2x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+3x& =5+1+2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+x-6& =0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Solve ${x}^{2}+x-6=0$ using the quadratic formula: $\begin{array}{llll}\hfill x& =\frac{-1±\sqrt{{1}^{2}-4\cdot 1\cdot \left(-6\right)}}{2\cdot 1},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-1±\sqrt{1+24}}{2},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-1±\sqrt{25}}{2},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-1±5}{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

That makes: $\begin{array}{llll}\hfill {x}_{1}& =\frac{-1+5}{2}=\frac{4}{2}=2,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}& =\frac{-1-5}{2}=\frac{-6}{2}=-3.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

3.
Put the solutions into (1) and calculate one $y$-value for ${x}_{1}$ and one $y$-value for ${x}_{2}$: $\begin{array}{l}\hfill \end{array}$
y1 = 1 2 2 = 1 4 = 5 y2 = 1 2 (3) = 1 + 6 = 5
y1 = 1 2 2 y2 = 1 2 (3) = 1 4 = 1 + 6 = 5 = 5

Solutions: $\begin{array}{llll}\hfill \left({x}_{1},{y}_{1}\right)& =\left(2,-5\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left({x}_{2},{y}_{2}\right)& =\left(-3,5\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Note! It is important to notice that an answer consists of one $x$-value and one $y$-value together! In this particular case, there are two answers.