$ \left\{ \begin{array}{l}{{x}^{2}}+{{y}^{2}}=61\\y-x=1\end{array} \right.$

Check the answers after solving:

$ \begin{align}{{\left( {-6} \right)}^{2}}+{{\left( {-5} \right)}^{2}}&=61\,\,\,\surd \\\left( {-5} \right)-\left( {-6} \right)&=1\,\,\,\,\,\,\surd \\{{\left( 5 \right)}^{2}}+{{\left( 6 \right)}^{2}}&=61\,\,\,\surd \\6-5&=1\,\,\,\,\,\,\surd \end{align}$

$ \begin{array}{c}y=x+1\\{{x}^{2}}+{{\left( {x+1} \right)}^{2}}=61\\{{x}^{2}}+{{x}^{2}}+2x+1=61\\2{{x}^{2}}+2x-60=0\\{{x}^{2}}+x-30=0\end{array}$

$ \begin{array}{c}{{x}^{2}}+x-30=0\\\left( {x+6} \right)\left( {x-5} \right)=0\\x=-6\,\,\,\,\,\,\,\,\,x=5\\y=-6+1=-5\,\,\,\,\,y=5+1=6\end{array}$

Answers are: $ \left( {-6,-5} \right)$ and $ \left( {5,6} \right)$

$ \left\{ \begin{array}{l}{{x}^{2}}+{{y}^{2}}=41\\xy=20\end{array} \right.$

Check the answers after solving:

$ \displaystyle \begin{array}{c}{{\left( 4 \right)}^{2}}+{{\left( 5 \right)}^{2}}=41\,\,\,\surd ;\,{{\left( {-5} \right)}^{2}}+{{\left( {-4} \right)}^{2}}=41\,\,\,\surd \\{{\left( 5 \right)}^{2}}+{{\left( 4 \right)}^{2}}=41\,\,\,\surd ;\,{{\left( {-5} \right)}^{2}}+{{\left( {-4} \right)}^{2}}=41\,\,\,\surd \\\left( 4 \right)\left( 5 \right)=20\,\,\,\surd ;\,\,\left( {-4} \right)\left( {-5} \right)=20\,\,\,\surd \\\left( 5 \right)\left( 4 \right)=20\,\,\,\surd ;\,\,\left( {-5} \right)\left( {-4} \right)=20\,\,\,\surd \,\,\,\,\,\,\end{array}$

$ \displaystyle y=\frac{{20}}{x};\,\,\,\,{{x}^{2}}+{{\left( {\frac{{20}}{x}} \right)}^{2}}=41$

$ \displaystyle {{x}^{2}}\left( {{{x}^{2}}+\frac{{400}}{{{{x}^{2}}}}} \right)=41{{x}^{2}}$

$ \displaystyle \begin{array}{c}{{x}^{4}}+400=41{{x}^{2}}\\\,{{x}^{4}}-41{{x}^{2}}+400=0\end{array}$

$ \begin{array}{c}{{x}^{4}}-41{{x}^{2}}+400=0\\\left( {{{x}^{2}}-16} \right)\left( {{{x}^{2}}-25} \right)=0\\{{x}^{2}}-16=0;\,\,\,x=\pm 4\,\,\,\,\,{{x}^{2}}-25=0;\,\,\,x=\pm 5\end{array}$

For $ x=4$: $ y=5$      $ x=5$: $ y=4$

$ x=-4$: $ y=-5$       $ x=-5$: $ y=-4$

Answers are: $ \left( {4,5} \right),\left( {-4,-5} \right),\left( {5,4} \right),$ and $ \left( {-5,-4} \right)$

$ \left\{ \begin{array}{l}4{{x}^{2}}+{{y}^{2}}=25\\3{{x}^{2}}-5{{y}^{2}}=-33\end{array} \right.$

Check the answers after solving:

$ \displaystyle \begin{array}{c}4{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}=25\surd \,;\,\,4{{\left( 2 \right)}^{2}}+{{\left( {-3} \right)}^{2}}=25\surd \\4{{\left( {-2} \right)}^{2}}+{{\left( 3 \right)}^{2}}=25\surd ;\,\,4{{\left( {-2} \right)}^{2}}+{{\left( {-3} \right)}^{2}}=25\surd \\3{{\left( 2 \right)}^{2}}-5{{\left( 3 \right)}^{2}}=-33\surd ;\,\,3{{\left( 2 \right)}^{2}}-5{{\left( {-3} \right)}^{2}}=-33\surd \\3{{\left( {-2} \right)}^{2}}-5{{\left( 3 \right)}^{2}}=-33\surd \,;\,\,3{{\left( {-2} \right)}^{2}}-5{{\left( {-3} \right)}^{2}}=-33\surd \end{array}$

$ \displaystyle \begin{array}{l}5\left( {4{{x}^{2}}+{{y}^{2}}} \right)=5\left( {25} \right)\\\,\,\,20{{x}^{2}}+5{{y}^{2}}=\,125\\\,\,\underline{{\,\,\,3{{x}^{2}}-5{{y}^{2}}=-33}}\\\,\,\,\,23{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,=92\\\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\pm 2\end{array}$

$ \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-2:\\4{{\left( 2 \right)}^{2}}+{{y}^{2}}=25\,\,\,\,\,\,\,\,4{{\left( -2 \right)}^{2}}+{{y}^{2}}=25\\{{y}^{2}}=25-16=9\,\,\,\,\,{{y}^{2}}=25-16=9\\\,\,\,\,\,\,\,\,\,y=\pm 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\pm 3\end{array}$

Answers are: $ \left( {2,3} \right),\,\,\left( {2,-3} \right),\,\,\left( {-2,3} \right),$ and $ \left( {-2,-3} \right)$

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

Check the answers after solving:

$ \displaystyle \begin{array}{c}-2={{\left( {-2} \right)}^{3}}-2{{\left( {-2} \right)}^{2}}-3\left( {-2} \right)+8\,\,\surd \\-2=-8-8+6+8\,\,\,\surd \,\end{array}$

$ \begin{array}{c}x={{x}^{3}}-2{{x}^{2}}-3x+8\\{{x}^{3}}-2{{x}^{2}}-4x+8=0\\{{x}^{2}}\left( {x-2} \right)-4\left( {x-2} \right)=0\\\left( {{{x}^{2}}-4} \right)\left( {x-2} \right)=0\\x=\pm 2\end{array}$

Answers are: $ (-2,-2)$ and $ (2,2)$

$ \left\{ \begin{array}{l}{{x}^{2}}+xy=4\\{{x}^{2}}+2xy=-28\end{array} \right.$

Check the answers after solving:

$ \displaystyle \begin{array}{c}{{\left( 6 \right)}^{2}}+\,\,\left( 6 \right)\left( {-\frac{{16}}{3}} \right)=4\,\,\,\surd \\{{\left( {-6} \right)}^{2}}+\,\,\left( {-6} \right)\left( {\frac{{16}}{3}} \right)=4\,\,\,\surd \\{{6}^{2}}+2\left( 6 \right)\left( {-\frac{{16}}{3}} \right)=-28\,\,\,\surd \\{{\left( {-6} \right)}^{2}}+2\left( {-6} \right)\left( {\frac{{16}}{3}} \right)=-28\,\,\,\surd \end{array}$

$ \displaystyle y=\frac{{4-{{x}^{2}}}}{x}$

$ \displaystyle \require {cancel} {{x}^{2}}+2\cancel{x}\left( {\frac{{4-{{x}^{2}}}}{{\cancel{x}}}} \right)=-28$

$ \displaystyle \begin{array}{c}{{x}^{2}}+8-2{{x}^{2}}=-28\\-{{x}^{2}}=-36\\x=\pm 6\end{array}$

$ x=6:\,\,\,\,\,\,\,\,\,\,\,\,x=-6:$

$ \displaystyle y=\frac{{4-{{6}^{2}}}}{6}\,\,\,\,\,\,\,\,\,y=\frac{{4-{{{\left( {-6} \right)}}^{2}}}}{{-6}}$

$ \displaystyle y=-\frac{{16}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=\frac{{16}}{3}$

Answers are: $ \displaystyle \left( {6,-\frac{{16}}{3}} \right)$ and $ \displaystyle \left( {-6,\frac{{16}}{3}} \right)$

$ \displaystyle \begin{array}{c}y={{e}^{x}}\\y-4{{x}^{2}}+1=0\end{array}$

Put in graphing calculator:

$ \displaystyle \begin{align}{{Y}_{1}}&={{e}^{x}}\\{{Y}_{2}}&=4{{x}^{2}}-1\end{align}$

Use the intersect feature on the calculator (2^nd trace, 5, enter, enter, enter) to find the intersection. (Use trace and arrow keys to get close to each intersection before using intersect). There are three solutions: $ \left( {-.62,.538} \right)$, $ \left( {.945,2.57} \right)$ and $ \left( {4.281,72.303} \right)$.

$ \begin{align}{{x}^{3}}+{{\left( {x-3} \right)}^{3}}&=407\\{{x}^{3}}+\left( {x-3} \right)\left( {{{x}^{2}}-6x+9} \right)&=407\\{{x}^{3}}+{{x}^{3}}-6{{x}^{2}}+9x-3{{x}^{2}}+18x-27&=407\\2{{x}^{3}}-9{{x}^{2}}+27x-434&=0\end{align}$

Use Synthetic Division (use $ 7$ as the first root; I found this root on the graphing calculator):

\begin{array}{l}\left. {\underline {\,
{\,\,7\,\,} \,}}\! \right| \,\,\,\,\,2\,\,-9\,\,\,\,\,\,27\,\,-434\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,35\,\,\,\,\,\,\,\,434\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,62\,\,\,\,\,\,\,\,\left| \!{\overline {\,
{\,\,0\,\,} \,}} \right. \end{array}

$ 2{{x}^{2}}+5x+62$ is prime (can’t be factored for real numbers), so the only root is $ 7$. $ x=7$ works, and to find $ y$, use $ y=x-3$. When $ x=7,\,\,y=4$. The two numbers are $ 4$ and $ 7$. They work!

The distance that Lacy has traveled in feet after $ t$ seconds can be modeled by the equation $ d\left( t\right)=150+75t-1.2{{t}^{2}}$. The distance that the police car travels after $ t$ seconds can be modeled by the equation $ d\left( t \right)=4{{t}^{2}}$.

(a)  How long will it take the police car to catch up to Lacy? 

(b)  How many feet has Lacy traveled from the time she saw the police car (time $ t=0$) until the police car catches up to Lacy?

(a) We can solve the systems of equations, using substitution by just setting the $ d\left( t \right)$’s ($ y$’s) together; use the Quadratic Formula to get the values for $ t$:

$ \displaystyle \begin{array}{c}150+75t-1.2{{t}^{2}}=4{{t}^{2}}\\5.2{{t}^{2}}-75t-150=0\end{array}$

$ \displaystyle t=\frac{{-\left( {-75} \right)\pm \sqrt{{{{{\left( {-75} \right)}}^{2}}-4\left( {5.2} \right)\left( {-150} \right)}}}}{{2\left( {5.2} \right)}}$

$ \displaystyle t\approx 16.20$

Note that we only want the positive value for $ t$, so in about $ 16.2$ seconds, the police car will catch up with Lacy.

(b) Plug the $ x$-value ($ t$) into either equation to get the $ y$-value ($ d(t)$); it’s easiest to use the second equation: $ d\left( t \right)=4{{\left( {16.2} \right)}^{2}}\approx 1050$. Lacy will have traveled about $ 1050$ feet when the police car catches up to her.

Note: we could have also solved using a graphing calculator: