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Could we have substituted values for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into the second equation to solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in [link] ?
Yes, but because $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is squared in the second equation this could give us extraneous solutions for $\text{\hspace{0.17em}}x.$
For $\text{\hspace{0.17em}}y=1$
This gives us the same value as in the solution.
For $\text{\hspace{0.17em}}y=2$
Notice that $\text{\hspace{0.17em}}\mathrm{-1}\text{\hspace{0.17em}}$ is an extraneous solution.
Solve the given system of equations by substitution.
$\left(-\frac{1}{2},\frac{1}{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,8\right)$
Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
[link] illustrates possible solution sets for a system of equations involving a circle and a line.
Given a system of equations containing a line and a circle, find the solution.
Find the intersection of the given circle and the given line by substitution.
One of the equations has already been solved for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ We will substitute $\text{\hspace{0.17em}}y=3x\mathrm{-5}\text{\hspace{0.17em}}$ into the equation for the circle.
Now, we factor and solve for $\text{\hspace{0.17em}}x.$
Substitute the two x -values into the original linear equation to solve for $\text{\hspace{0.17em}}y.$
The line intersects the circle at $\text{\hspace{0.17em}}\left(2,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,\mathrm{-2}\right),$ which can be verified by substituting these $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ values into both of the original equations. See [link] .
Solve the system of nonlinear equations.
$\left(\mathrm{-1},3\right)$
We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.
[link] illustrates possible solution sets for a system of equations involving a circle and an ellipse .
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