Chapter 1, Problem 3CT

To determine

To solve: The given system of linear equations and then to verify the answer.

Answer to Problem 3CT

The solution is $(26,-3,15)$ .

Explanation of Solution

Given information: A system of linear equations.

Calculation:

  $\begin{array}{l}-2x+y+4z=5-(i)\\ x+3y-z=2-(ii)\\ 4x+y-6z=11-(iii)\end{array}$

Multiplying $(ii)$ by $2$ and adding it to $(i)$ ,

  $\begin{array}{l}2x+6y-2z=4\\ -2x+y+4z=5\\ ==============\\ 7y+2z=9\text{}-(iv)\end{array}$

Multiplying $(ii)$ by $4$ and subtracting $(iii)$ from it,

  $\begin{array}{l}4x+12y-4z=8\\ 4x+y-6z=11\\ =============\\ 11y+2z=-3-(v)\end{array}$

Subtracting $(iv)$ from $(v)$ ,

  $\begin{array}{l}11y+2z=-3\\ 7y+2z=9\\ ==========\\ 4y=-12\\ y=-3\end{array}$

Substituting derived value $y$ of in $(iv)$ ,

  $\begin{array}{l}7(-3)+2z=9\\ 2z=30\\ z=15\end{array}$

Substituting derived values of $z$ and $y$ in $(ii)$ ,

  $\begin{array}{l}x+3(-3)-15=2\\ x=26\end{array}$

Verifying by putting values of $x,yandz$ in $(i),(ii)and(iii)$ ,

  $\begin{array}{l}-2(26)-3+4(15)=5\\ 5+60=5\\ 5=5\\ L.H.S.=R.H.S.\\ 26+3(-3)-15=2\\ 26-9-15=2\\ 2=2\\ L.H.S.=R.H.S.\\ 4(26)-3-6(15)=11\\ 104-3-90=11\\ 11=11\\ L.H.S.=R.H.S.\end{array}$

Hence, verified.

Thus, the solution is $(26,-3,15)$ .