# 7.3 Systems of nonlinear equations and inequalities: two variables  (Page 2/9)

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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 $\text{\hspace{0.17em}}y=1$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}{x}^{2}=0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±\sqrt{0}=0\hfill \end{array}$

This gives us the same value as in the solution.

For $\text{\hspace{0.17em}}y=2$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={x}^{2}+1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2={x}^{2}+1\hfill \\ {x}^{2}=1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±\sqrt{1}=±1\hfill \end{array}$

Notice that $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ is an extraneous solution.

Solve the given system of equations by substitution.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x-y=-2\hfill \\ 2{x}^{2}-y=0\hfill \end{array}$

$\left(-\frac{1}{2},\frac{1}{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,8\right)$

## Intersection of a circle and a line

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.

## Possible types of solutions for the points of intersection of a circle and a line

[link] illustrates possible solution sets for a system of equations involving a circle and a line.

• No solution. The line does not intersect the circle.
• One solution. The line is tangent to the circle and intersects the circle at exactly one point.
• Two solutions. The line crosses the circle and intersects it at two points.

Given a system of equations containing a line and a circle, find the solution.

1. Solve the linear equation for one of the variables.
2. Substitute the expression obtained in step one into the equation for the circle.
3. Solve for the remaining variable.
4. Check your solutions in both equations.

## Finding the intersection of a circle and a line by substitution

Find the intersection of the given circle and the given line by substitution.

$\begin{array}{l}{x}^{2}+{y}^{2}=5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=3x-5\hfill \end{array}$

One of the equations has already been solved for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ We will substitute $\text{\hspace{0.17em}}y=3x-5\text{\hspace{0.17em}}$ into the equation for the circle.

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+{\left(3x-5\right)}^{2}=5\\ {x}^{2}+9{x}^{2}-30x+25=5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}10{x}^{2}-30x+20=0\end{array}$

Now, we factor and solve for $\text{\hspace{0.17em}}x.$

$\begin{array}{l}\text{\hspace{0.17em}}10\left({x}^{2}-3x+2\right)=0\hfill \\ 10\left(x-2\right)\left(x-1\right)=0\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=2\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1\hfill \end{array}$

Substitute the two x -values into the original linear equation to solve for $\text{\hspace{0.17em}}y.$

$\begin{array}{l}y=3\left(2\right)-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\hfill \\ \hfill \\ \hfill \\ \hfill \\ y=3\left(1\right)-5\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\hfill \end{array}$

The line intersects the circle at $\text{\hspace{0.17em}}\left(2,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,-2\right),$ 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.

$\begin{array}{l}{x}^{2}+{y}^{2}=10\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-3y=-10\hfill \end{array}$

$\left(-1,3\right)$

## Solving a system of nonlinear equations using elimination

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.

## Possible types of solutions for the points of intersection of a circle and an ellipse

[link] illustrates possible solution sets for a system of equations involving a circle and an ellipse .

• No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
• One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point.
• Two solutions. The circle and the ellipse intersect at two points.
• Three solutions. The circle and the ellipse intersect at three points.
• Four solutions. The circle and the ellipse intersect at four points.

12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
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Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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Abhi
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salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8