Chapter 11, Problem 16RE

To determine

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

Answer to Problem 16RE

  $\boxed{(x=-3,y=3,z=10)}$

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.

  $\{\begin{array}{l}x+5y-z=2\\ 2x-y+z=7\\ x-y+2z=11\end{array}$

Calculation:

The given system of equations in three variables is given by the expressions given below.

  $\{\begin{array}{l}x+5y-z=2\\ 2x-y+z=7\\ x-y+2z=11\end{array}$

The given system consists of three variables from which one of the variables is required to be removed from the equations. This can be done by multiplying first equations by $-2$ and the second equation by $2$ .the equations obtained are given below.

  $\{\begin{array}{l}-2x-10y+2z=-4\\ 4x+2y+2x=14\\ x-y+2z=11\end{array}$

We can now obtain two equations in $x$ and $y$ which can be solved simultaneously by subtracting

Second equation from the first equation and by subtracting third equation from the first equation.

  $\{\begin{array}{l}4x-(-2x)+2y-(-10y)+2z-2z=14-(-4)\\ x-(-2x)-y-(-10y)+2z-3z=11-(-4)\end{array}$

  $\{\begin{array}{l}6x+12y=18\\ 3x+9y=15\end{array}$

The given system of equations can be solved by the method of elimination. The first equation can be multiplied by 1 and second equation can be multiplied by 2 in order make the coefficient of $x$ same in both the equations.

  $\{\begin{array}{l}(6x+12y=18)\times 1\\ (3x+9y=15)\times 2\end{array}$

  $\{\begin{array}{l}6x+12y=18\\ 6x+18y=30\end{array}$

The equation can now be solved for $y$ since the coefficients of $x$ is same in both the equations

  $6x+18y-6x-12y=30-12$

  $6y=18$

  $y=3$

The value of $x$ can be determined by substituting the obtained value of $y$ .

  $6x+12(3)=18$

  $6x=18-36$

  $x=\frac{-18}{6}$$x=-3$

The value of $z$ can be obtained by substituting the value of $x$ and $y$ in any of the given equations.

  $x+5y-z=2$

  $-3+5(3)-z=2$

  $-z=2-15+3$

  $z=10$

Hence, the solution of the given system of equation is $\boxed{(x=-3,y=3,z=10)}$ .