7.3 Systems of nonlinear equations and inequalities: two variables  (Page 3/9)

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Solving a system of nonlinear equations representing a circle and an ellipse

Solve the system of nonlinear equations.

$\begin{array}{rr}\hfill {x}^{2}+{y}^{2}=26\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \hfill \left(1\right)\\ \hfill 3{x}^{2}+25{y}^{2}=100& \hfill \left(2\right)\end{array}$

Let’s begin by multiplying equation (1) by $\text{\hspace{0.17em}}-3,\text{}$ and adding it to equation (2).

After we add the two equations together, we solve for $\text{\hspace{0.17em}}y.$

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

Substitute $\text{\hspace{0.17em}}y=±1\text{\hspace{0.17em}}$ into one of the equations and solve for $\text{\hspace{0.17em}}x.$

There are four solutions: $\text{\hspace{0.17em}}\left(5,1\right),\left(-5,1\right),\left(5,-1\right),\text{and}\text{\hspace{0.17em}}\left(-5,-1\right).\text{\hspace{0.17em}}$ See [link] .

Find the solution set for the given system of nonlinear equations.

$\begin{array}{c}4{x}^{2}+{y}^{2}=13\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=10\end{array}$

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

Graphing a nonlinear inequality

All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality    is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality.

Recall that when the inequality is greater than, $\text{\hspace{0.17em}}y>a,$ or less than, $\text{\hspace{0.17em}}y the graph is drawn with a dashed line. When the inequality is greater than or equal to, $\text{\hspace{0.17em}}y\ge a,\text{}$ or less than or equal to, $\text{\hspace{0.17em}}y\le a,\text{}$ the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See [link] .

Given an inequality bounded by a parabola, sketch a graph.

1. Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set.
2. If the boundary is included in the region (the operator is $\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\ge$ ), the parabola is graphed as a solid line.
3. If the boundary is not included in the region (the operator is<or>), the parabola is graphed as a dashed line.
4. Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line.
5. Shade the region representing the solution set.

Graphing an inequality for a parabola

Graph the inequality $\text{\hspace{0.17em}}y>{x}^{2}+1.$

First, graph the corresponding equation $\text{\hspace{0.17em}}y={x}^{2}+1.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}y>{x}^{2}+1\text{\hspace{0.17em}}$ has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points
$\left(0,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(2,0\right).\text{\hspace{0.17em}}$ One point is clearly inside the parabola and the other point is clearly outside.

$\begin{array}{ll}y>{x}^{2}+1\hfill & \hfill \\ 2>{\left(0\right)}^{2}+1\hfill & \hfill \\ 2>1\hfill & \text{True}\hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ \hfill & \hfill \\ 0>{\left(2\right)}^{2}+1\hfill & \hfill \\ 0>5\hfill & \text{False}\hfill \end{array}$

The graph is shown in [link] . We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

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?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
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
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali