11.3 Systems of nonlinear equations and inequalities: two variables  (Page 5/9)

Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers.

When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution?

When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?

If you graph a revenue and cost function, explain how to determine in what regions there is profit.

If you perform your break-even analysis and there is more than one solution, explain how you would determine which x -values are profit and which are not.

$\begin{array}{l}x+y=4\hfill \\ x^2+y^2=9\hfill \end{array}$

$\begin{array}{l}y=x\mathrm{-3}\hfill \\ x^2+y^2=9\hfill \end{array}$

$\begin{array}{l}y=x\hfill \\ x^2+y^2=9\hfill \end{array}$

$\begin{array}{l}y=-x\hfill \\ x^2+y^2=9\hfill \end{array}$

$\begin{array}{l}x=2\hfill \\ x^2-y^2=9\hfill \end{array}$

$\begin{array}{l}\hfill \\ 4x^2\mathrm{-9}{y}^2=36\hfill \\ 4x^2+9y^2=36\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=25\\ x^2-y^2=1\end{array}$

$\begin{array}{l}\hfill \\ 2x^2+4y^2=4\hfill \\ 2x^2\mathrm{-4}{y}^2=25x\mathrm{-10}\hfill \end{array}$

$\begin{array}{l}{y}^2-x^2=9\\ 3x^2+2y^2=8\end{array}$

$\begin{array}{l}{x}^2+y^2+\frac{1}{16}=2500\\ y=2x^2\end{array}$

$\begin{array}{l}\begin{array}{l}\\ \mathrm{-2}{x}^2+y=\mathrm{-5}\end{array}\hfill \\ 6x-y=9\hfill \end{array}$

$\begin{array}{l}\begin{array}{l}\\ -x^2+y=2\end{array}\hfill \\ -x+y=2\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=1\hfill \\ y=20x^2\mathrm{-1}\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=1\hfill \\ y=-x^2\hfill \end{array}$

$\begin{array}{l}2x^3-x^2=y\hfill \\ y=\frac{1}{2}-x\hfill \end{array}$

$\begin{array}{l}9x^2+25y^2=225\hfill \\ {(x\mathrm{-6})}^2+y^2=1\hfill \end{array}$

$\begin{array}{l}{x}^4-x^2=y\hfill \\ x^2+y=0\hfill \end{array}$

$\begin{array}{l}2x^3-x^2=y\hfill \\ x^2+y=0\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=9\hfill \\ y=3-x^2\hfill \end{array}$

$\begin{array}{l}{x}^2-y^2=9\hfill \\ x=3\hfill \end{array}$

$\begin{array}{l}{x}^2-y^2=9\hfill \\ y=3\hfill \end{array}$

$\begin{array}{l}{x}^2-y^2=9\hfill \\ x-y=0\hfill \end{array}$

$\begin{array}{l}\hfill \\ -x^2+y=2\hfill \\ \mathrm{-4}x+y=\mathrm{-1}\hfill \end{array}$

$\begin{array}{l}-x^2+y=2\hfill \\ 2y=-x\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=25\\ x^2-y^2=36\end{array}$

$\begin{array}{l}{x}^2+y^2=1\hfill \\ y^2=x^2\hfill \end{array}$

$\begin{array}{l}16x^2\mathrm{-9}{y}^2+144=0\hfill \\ y^2+x^2=16\hfill \end{array}$

$\begin{array}{l}3x^2-y^2=12\hfill \\ {(x\mathrm{-1})}^2+y^2=1\hfill \end{array}$

$\begin{array}{l}3x^2-y^2=12\hfill \\ {(x\mathrm{-1})}^2+y^2=4\hfill \end{array}$

$\begin{array}{l}3x^2-y^2=12\hfill \\ x^2+y^2=16\hfill \end{array}$

$\begin{array}{l}{x}^2-y^2-6x-4y-11=0\hfill \\ -x^2+y^2=5\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2\mathrm{-6}y=7\hfill \\ x^2+y=1\hfill \end{array}$

$\begin{array}{l}{x}^2+y^2=6\hfill \\ xy=1\hfill \end{array}$

$x^2+y<9$

$x^2+y^2<4$

$\begin{array}{l}{x}^2+y<1\\ y>2x\end{array}$

$\begin{array}{l}{x}^2+y<\mathrm{-5}\\ y>5x+10\end{array}$

$\begin{array}{l}{x}^2+y^2<25\\ 3x^2-y^2>12\end{array}$

$\begin{array}{l}{x}^2-y^2>\mathrm{-4}\\ x^2+y^2<12\end{array}$

$\begin{array}{l}{x}^2+3y^2>16\\ 3x^2-y^2<1\end{array}$

$\begin{array}{l}\hfill \\ y\ge e^x\hfill \\ y\le \ln (x)+5\hfill \end{array}$

$\begin{array}{l}y\le -\log (x)\\ y\le e^x\end{array}$

$\begin{array}{l}\frac{4}{x^2}+\frac{1}{y^2}=24\\ \frac{5}{x^2}-\frac{2}{y^2}+4=0\end{array}$

$\begin{array}{c}\frac{6}{x^2}-\frac{1}{y^2}=8\\ \frac{1}{x^2}-\frac{6}{y^2}=\frac{1}{8}\end{array}$

$\begin{array}{l}{x}^2-xy+y^2\mathrm{-2}=0\hfill \\ x+3y=4\hfill \end{array}$

$\begin{array}{l}{x}^2-xy\mathrm{-2}{y}^2\mathrm{-6}=0\hfill \\ x^2+y^2=1\hfill \end{array}$

$\begin{array}{l}{x}^2+4xy\mathrm{-2}{y}^2\mathrm{-6}=0\hfill \\ x=y+2\hfill \end{array}$

$\begin{array}{l}xy<1\\ y>\sqrt{x}\end{array}$

$\begin{array}{l}{x}^2+y<3\\ y>2x\end{array}$

Two numbers add up to 300. One number is twice the square of the other number. What are the numbers?

The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers?

A laptop company has discovered their cost and revenue functions for each day: $C(x)=3x^2\mathrm{-10}x+200$ and $R(x)=\mathrm{-2}{x}^2+100x+50.$ If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit.

A cell phone company has the following cost and revenue functions: $C(x)=8x^2\mathrm{-600}x+\mathrm{21,500}$ and $R(x)=\mathrm{-3}{x}^2+480x.$ What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit.