Chapter 3.4, Problem 11PPS

To determine

To calculate: each system of given equation and find the value of $r,s$ and $t$ .

Answer to Problem 11PPS

The value of $r,s$ and $t$ are: $-2,-1\text{ and }4$ .

Explanation of Solution

Given information:

The system of equation: $\begin{array}{l}4r+6s-t=-18\hfill \\ 3r+2s-4t=-24\hfill \\ -5r+3s+2t=15\hfill \end{array}$

Calculation:

Let, the equations are,

  $\begin{array}{l}4r+6s-t=-\mathrm{18.......}\left(1\right)\hfill \\ 3r+2s-4t=-24\dots \dots (2)\hfill \\ -5r+3s+2t=15\dots \dots (3)\hfill \end{array}$

To solve the above equation by eliminate one variable using the two equations. First take the equations (1) and (2)

  $4r+6s-t=-18$

  $3r+2s-4t=-24$

Multiply by $4$ in equation (2) and 3 in equation (1) and simplify it as follows,

  $\begin{array}{l}12r+18s-3t=-54\\ \frac{(-)12r+8s-16t=-96}{10s+13t=42}\end{array}$

Therefore $10s+13t=42$ is considered as equation (4).

Take the given equations (2) and equation $(3),$

  $\begin{array}{l}3r+2s-4t=-24\\ -5r+3s+2t=15\end{array}$

Multiply by 3 in equation (3) and 5 in equation (2) and simplify it as follows,

  $\begin{array}{l}15r+10s-20t=-120\\ \frac{(-)15r+9s+6t=45}{19s-14t=-75}\end{array}$

Therefore, $19s-14t=-75$ is considered as equation (5).

Take equation (4) and (5)

  $10s+13t=42$

  $19s-14t=-75$

Multiply by 26 in equation (5) and 14 in equation (4) and $\text{simplif}y$ it as follows,

  $190s+247t=798$

  $\frac{(-)190s-140t=-750}{387t=1548}$

  $t=\frac{{1548}^4}{{387}_1}$

  $t=4$

Substitute the value of $t=4$ in equation (4) and simplify it as follows,

  $\begin{array}{l}10s+13(4)=42\\ 10s+52=42\\ 10s=42-52\\ s=\frac{-10}{10}\end{array}$

  $s=-1$

Substitute the value of $s=-1$ , $t=4$ in equation (1) and simplify it as follows, $4r+6(-1)-(4)=-18$

  $\begin{array}{r}\hfill 4r-6-4=-18\\ \hfill 4r-10=-18\\ \hfill 4r=-18+10\\ \hfill r=\frac{-8^2}{4_1}\\ \hfill r=-2\end{array}$

Hence, the value of $r,s$ and $t$ are: $-2,-1\text{ and }4$ .