Chapter 7.3, Problem 35E

To determine

To Solve : System of linear equation

Answer to Problem 35E

The given System of equation has infinitely many solution

  $z=a$ , $x=-2a-1$ , $y=\frac{1}{2}\left(a+2\right)$

Explanation of Solution

Given information:

The equation is

  $\begin{array}{l}x+2y+z=1\\ x-2y+3z=-3\\ 2x+y+z=-1\end{array}$

Calculation :

The system of linear equation is

  $\begin{array}{l}x+2y+z=1\\ x-2y+3z=-3\\ 2x+y+z=-1\end{array}$

Using Elimination method

Equation 1: $x+2y+z=1$

Equation 2: $x-2y+3z=-3$

Equation 3: $2x+y+z=-1$

Now , Subtract Equation 1 and Equation 2 to obtain Equation 4

  $x+2y+z=1$

  $x-2y+3z=-3$

  $2x+4z=-2$

Equation 4: $2x+4z=-2$

Now ,Multiply (Equation 3) by 2 and Add it with (Equation 2)

  $x-2y+3z=-3$

  $4x+2y+2z=-2$

  $5x+5z=-5$

  $x+z=-1$

Equation 5: $x+z=-1$

Now , Multiply (Equation 5) by 2 and Subtract it with (Equation 4)

  $2x+4z=-2$

  $2x+4z=-2$

  $0+0=0$

This means that Equation 3 depends on Equation 1 and Equation 2 , it does not give any additional information therefore given system of equation is equivalent to

  $x+2y+z=1$

  $x-2y+3z=-3$ and has infinitely many solution

Let $z=a$ Where $a$ is a real number

Therefore $x=1-2y-a$

Putting Value of $x$ in Equation 2

  $1-2y-a-2y+3a=-3$

So, $y=\frac{1}{2}\left(a+2\right)$

Putting value of $y$ in $x=1-2y-a$

So, $x=-2a-1$

Hence the given System of equation has infinitely many solution

  $z=a$ , $x=-2a-1$ , $y=\frac{1}{2}\left(a+2\right)$