Chapter 7.1, Problem 50E

To determine

The solution of the system algebraically and graphically.

Answer to Problem 50E

  $(3,4)(5,0)$ .

Explanation of Solution

Given information:

The system of the equation are:

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

Calculation:

The given equations are $\{\begin{array}{l}{x}^2+y^2=25\\ 2x+y=10\end{array}$ .

From the second equation isolate the value of $x$ in terms of $y$ :

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

Substitute the value of $x$ in equation first and find the value of $y$ :

  $\begin{array}{l}{x}^2\text{+}{y}^2\text{=25 }\\ {\left(\frac{10-y}{2}\right)}^2+y^2=25\\ 100+5y^2-20y=100\\ 5y^2-20y=0\text{ {Factorise the equation}}\\ 5y(y-4)=0\\ y=4,0\end{array}$

Substitute the value of $y$ in equation $x=\frac{10-y}{2}$ are as shown below:

  $\begin{array}{l}\text{For }y=4\\ x=\frac{10-4}{2}\\ x=3\\ \text{For }y=0\\ x=\frac{10-0}{2}\\ x=5\end{array}$

The ordered pairs are $(3,4)(5,0)$ .

Now graph the above system of equations graphically are as follows:

  

The lines are intersected at $(3,4)(5,0)$ .

Thus, the solution for the system of equations is the ordered pair $(3,4)(5,0)$ .