Chapter 7, Problem 8CR

To determine

To solve: The system by method of substitution.

Answer to Problem 8CR

The solution of the system is $\left\{(13,0),(5,12)\right\}$ .

Explanation of Solution

Given information: The given system is:

  $\begin{array}{l}{x}^2+y^2=169(i)\\ 3x+2y=39(ii)\end{array}$ .

Calculation: The given system is:

  $\begin{array}{l}{x}^2+y^2=169(i)\\ 3x+2y=39(ii)\end{array}$ .

To solve the system by method of substitution:

From equation $(ii)$

  $\begin{array}{l}3x+2y=39\\ 2y=39-3x\\ y=\frac{39-3x}{2}(iii)\end{array}$

Put the value of $y=\frac{39-3x}{2}$ in equation $(i)$ :

  $\begin{array}{l}$ x^2+{(\frac{39-3x}{2})}^2=169$x^2+\frac{1521+9x^2-234x}{4}=169\\ \frac{4x^2+1521+9x^2-234x}{4}=169\\ 13x^2-234x+1521=169\times 4=676\\ 13x^2-234x+1521-676=0\\ 13x^2-234x+845=0\\ 13(x^2-18x+65)=0\\ x^2-13x-5x+65=0\\ x(x-13)-5(x-13)=0\\ (x-13)(x-5)=0\\ x-13=0,x-5=0\\ x=13,x=5\\ x=13,5\end{array}$

Put the value of $x=13\mathrm{and}5$ in equation $(iii)$ :

  $\begin{array}{l}\mathrm{put}x=13\\ y=\frac{39-3(13)}{2}=\frac{39-39}{2}=\frac{0}{2}=0\\ \mathrm{put}x=5\\ y=\frac{39-3(5)}{2}=\frac{39-15}{2}=\frac{24}{2}=12\end{array}$

Thus the solution of the system is $\left\{(13,0),(5,12)\right\}$ .