Chapter 6.4, Problem 1BGP

To determine

To solve the given system of equations

Answer to Problem 1BGP

  $r=2$ and $q=-5$

Explanation of Solution

Given:

Equations:

  $9r+q=13$

  $3r+2q=-4$

Calculation:

Let,

  $\begin{array}{l}9r+q=\mathrm{13.......}(i)\\ 3r+2q=-\mathrm{4.......}(ii)\end{array}$

Now, subtract equation ( i ) from equation $3\times$ ( ii ),

  $\begin{array}{l}\Rightarrow 3(3r+2q)-(9r+q)=3\times (-4)-13\\ \Rightarrow 9r+6q-9r-q=-25\\ \Rightarrow 5q=-25\\ \Rightarrow q=-5\end{array}$

Now, substitute the value of q in equation ( i ),

  $\begin{array}{l}\Rightarrow 9r+(-5)=13\\ \Rightarrow 9r=18\\ \Rightarrow r=2\end{array}$

Conclusion:

Therefore, the required solution for the given system of equation is $r=2$ and $q=-5.$