Chapter 11, Problem 17RE

To determine

Solve each system of equations using the method of substitution ot the method of elimination.

Answer to Problem 17RE

  $\left\{(x,y,z)|x=\frac{7}{4}z+\frac{39}{4},y=\frac{9}{8}z+\frac{69}{8},z,any-real-number\right\}$

Explanation of Solution

Given information:

Solve each system of equations using the method of substitution or the method of elimination. If the system has no solution, say that it is inconsistent.

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

Calculation:

We can use the elimination method to find the solutions of the system of equations.

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

So as to make the coefficients of $y$ negatives of one another.

  $2x+4y-8z=54$

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

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

So, we get an equation containing two variables $x$ and $z$ .

We need another equation containing the variables $x$ and $z$ to find their values.

Multiply both sides of the equation $x+2y-4z=27$ by $3$ so as to make the coefficients of $y$

negatives of one number.

  $3x+6y-12z=81$

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

  $\begin{array}{l}3x+6y-12z=81\\ \begin{array}{c}5x-6y-2z=-3\end{array}\\ \begin{array}{ccccc}8x& & -14z=78& & \end{array}\end{array}$

Divide both sides of the equation $8x-14z=78$ by $2$ .

Since, we get 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 $z$ .

  $4x=39+7z$

  $x=\frac{7}{4}z+\frac{39}{4}$

Substitute $\frac{7}{4}z+\frac{39}{4}$ for $x$ in the equation $x+2y-4z=27$ , to get the value of $y$ in terms of $z$ .

  $\frac{7}{4}z+\frac{39}{4}+2y-4z=27$

  $2y=4z-\frac{7}{4}z+27-\frac{39}{4}$

  $2y=\frac{9}{4}z+\frac{69}{4}$

  $y=\frac{9}{8}z+\frac{69}{8}$

Hence, the solution, in ordered triplet form, is

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