Chapter 10, Problem 10RE

To determine

To solve: the system of equations

Answer to Problem 10RE

The four solutions are $\left(1.41,2.83\right)and\left(-1.41,2.83\right)and\left(1,3.32\right)and\left(-1,3.32\right)$

Explanation of Solution

Given:

  $\{\begin{array}{l}{x}^2+y^2=10 \\ x^2+2y^2-7y=0 \end{array}$

Calculation:

Consider the following system of equations:

  $\{\begin{array}{l}{x}^2+y^2=10 \dots \dots \mathrm{..}\left(1\right)\\ x^2+2y^2-7y=0 \dots \dots \mathrm{..}\left(2\right)\end{array}$

Solve the equation $1for{x}^2$

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

Substitute:

Substitute $x^2=10-y^2$ in $x^2+2y^2-7y=0$ and solve for $y$ .

  $\begin{array}{l}{(10-y^2)}^2\text{ +2}{y}^2-7y=0 Substitute{x}^2=10-y^2\hfill \\ 100+y^4-20y^2+2y^2-7y=0 Expand\hfill \\ y^4-18y^2-7y+100=0 Combineliketerms\hfill \end{array}$

By trial and error method, get the real solutions.

  $y\text{ =}2.83ory=3.32$

By trial and error method, get the complex solutions.

  $y=-3.08+1.03iory=-3.08-1.03i$

Back-substitute:

Now substitute the values of $y$ into the equation $x^2+y^2=10$ .

  $\begin{array}{l} Fory=2.83:\hfill \\ x^2+{\left(2.83\right)}^2=10\hfill \\ x=\pm 1.41\hfill \\ Fory=3.32:\hfill \\ x^2+{\left(3.32\right)}^2=10\hfill \\ x=\pm 1\hfill \end{array}$

Thus, the four solutions are $\boxed{\left(1.41,2.83\right)and\left(-1.41,2.83\right)}and\boxed{\left(1,3.32\right)and\left(-1,3.32\right)}$

The following graph shows the intersecting points of the graphs at $\left(1.41,2.83\right)and\left(-1.41,2.83\right)$ and $\left(1,3.32\right)and\left(-1,3.32\right)$ .

  

Check:

Check the solution by substituting $\left(1.41,2.83\right)and\left(-1.41,2.83\right)$ in the system of equations.

  $Forx=1.41,y=2.83$

  $\{\begin{array}{l}{x}^2+y^2=10 \dots \dots .\left(1\right)\\ {\left(1.41\right)}^2+{\left(2.83\right)}^2=10\\ 10=10 true\end{array}$

  $Forx=-1.41,y=2.83$

  $\{\begin{array}{l}{x}^2+y^2=10 \dots \dots .\left(1\right)\\ {\left(-1.41\right)}^2+{\left(2.83\right)}^2=10\\ 10=10 true\end{array}$

Check the solution by substituting $\left(1,3.32\right)and\left(-1,3.32\right)$ in the system of equations.

  $Forx=1,y=3.32$

  $\{\begin{array}{l}{x}^2+y^2=10 \dots \dots .\left(1\right)\\ {\left(1\right)}^2+{\left(3.32\right)}^2=10\\ 10=10 true\end{array}$

  $Forx=1,y=-3.32$

  $\{\begin{array}{l}{x}^2+y^2=10 \dots \dots .\left(1\right)\\ {\left(1\right)}^2+{\left(-3.32\right)}^2=10\\ 10=10 true\end{array}$

Conclusion: The four solutions are $\left(1.41,2.83\right)and\left(-1.41,2.83\right)and\left(1,3.32\right)and\left(-1,3.32\right)$