Chapter 11, Problem 9RE

To determine

To find: The ordered triple form.

Answer to Problem 9RE

The triple product is $\left\{\left(x,y,z\right)|x=\frac{7}{4}z+\frac{39}{4},y=\frac{9}{8}z+\frac{69}{8},z\text{any}\text{real}\text{number}\right\}$ .

Explanation of Solution

Given information:

The given equations are,

  $\left\{\begin{array}{r}2x-4y+z=15\\ x-2y-4z=27\\ 5x-6y-2z=-3\end{array}\right\}$

Calculation:

Calculate the equations,

Multiply both sides of the equation $x+2y-4z=27$ by $2$ so as to make the coefficients is negative of the another.

  $2x+4y-8z=54$

Add the equation $2x+4y-8z=54$ and $2x+4y+z=-15$ .

  $\begin{array}{c}2x+4y-8z=54\\ 2x+4y+z=-15\\ 4x-7z=39\end{array}$

The equation containing two variables $x$ and $z$ .

Multiply both sides of the equation $x+2y-4z=27$ by $3$ .

  $3x+6y-12z=81$ .

Add the equation $3x+6y-12z=81$ and $5x-6y-2z=-3$ .

  $\begin{array}{c}3x+6y-12z=81\\ 5x-6y-2z=-3\\ 8x-17z=78\\ 4x-7z=39\end{array}$

The same equation in both cases, we cannot find the exact value of the variables $x$ and $z$ . So, express the value of $x$ in terms of $z$ .

  $\begin{array}{l}4x=39+7z\\ x=\frac{7}{4}z+\frac{39}{4}\end{array}$

Substitute the value in this equation $x+2y-4z=27$ ,

  $\begin{array}{c}\frac{7}{4}z+\frac{39}{4}+2y-4z=27\\ 2y=\frac{9}{4}z+\frac{69}{4}\\ y=\frac{9}{8}z+\frac{69}{8}\end{array}$

Therefore, the solution in ordered triple form is $\left\{\left(x,y,z\right)|x=\frac{7}{4}z+\frac{39}{4},y=\frac{9}{8}z+\frac{69}{8},zanyrealnumber\right\}$ .