Chapter 3.4, Problem 10PPS

To determine

To calculate: each system of given equation and find the value of $x,y$ and z.

Answer to Problem 10PPS

The value of $x,y$ and $z$ are: $8,-3\text{ and }2.$ .

Explanation of Solution

Given information:

The system of equation: $\begin{array}{l}-2x+5y+3z=-25\hfill \\ -4x-3y-8z=-39\hfill \\ 6x+8y-5z=14\hfill \end{array}$

Calculation:

Let, the equations are,

  $\begin{array}{l}-2x+5y+3z=-\mathrm{25.......}\left(1\right)\hfill \\ -4x-3y-8z=-39\dots \dots (2)\hfill \\ 6x+8y-5z=14\dots \dots (3)\hfill \end{array}$

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

  $-2x+5y+3z=-25$

  $-4x-3y-8z=-39$

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

  $\frac{(+)-8x-6y-16z=-78}{26y+28z=-22}$

Therefore $26y+28z=-22$ is considered as equation (4).

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

  $\begin{array}{cc}-4x-3y-8z& =-39\\ 6x+8y-5z& =14\end{array}$

Multiply by 4 in equation (3) and 6 in equation (2) and simplify it as follows,

  $\begin{array}{l}-24x-18y-48z=-234\\ \frac{(+)24x+32y-20z=56}{14y-68z=-178}\end{array}$

Therefore, $14y-68z=-178$ is considered as equation (5).

Take equation (4) and (5)

  $26y+28z=-22$

  $14y-68z=-178$

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

  $\begin{array}{l}364y+392z=-308\\ \frac{(-)364y-1768z=-4628}{2160z=4320}\end{array}$

  $z=\frac{{4320}^2}{{2160}_1}$

  $z=2$

Substitute the value of $z=2$ in equation (4) and simplify it as follows,

  $\begin{array}{l}26y+28(2)=-22\\ 26y+56=-22\\ 26y=-22-56\\ 26y=-78\end{array}$

  $y=\frac{-78}{26}$

  $y=-3$

Substitute the value of $y=-3$ and $z=2$ in equation (1) and simplify it as follows,

  $-2x+5(-3)+3(2)=-25$

  $\begin{array}{r}\hfill -2x-15+6=-25\\ \hfill -2x-9=-25\\ \hfill -2x=-25+9\\ \hfill -2x=-16\\ \hfill x=\frac{76}{72}\\ \hfill x=8\end{array}$

Hence, the value of $x,y$ and $z$ are: $8,-3\text{ and }2$ .