Chapter 7, Problem 43CR

To determine

To solve: System of linear equations.

Answer to Problem 43CR

  $x=\frac{-3}{4},y=0,z=\frac{-5}{4}$

Explanation of Solution

Given Information: - Given set of equations

  $2x+6z=-9,\text{ 3}x-2y+11z=-16,\text{ 3}x-y+7z=-11$

- The set of equations

  $2x+6z=-9(i)$

  $3x-2y+11z=-16(ii)$

  $3x-y+7z=-11(iii)$

Multiply eq (iii) by 2 and subtract form (ii)

  $\begin{array}{l}3x-2y+11z-2(3x-y+7z)=-16-2(-11)\\ 3x-2y+11z-6x+2y-142=-16+22\end{array}$

  $-3x-3z=6$

  $x+z=-2$ ----------------- (iv)

Multiply eq. (iv) by 2 and subtract it from (i)

  $\begin{array}{l}2x+6z-2(x+z)=-9-2(-2)\\ 2x+6z-2x-2z=-9+4\end{array}$

  $\begin{array}{l}4z=-5\\ z=\frac{-5}{4}\end{array}$

Put value of z in eq. (iv)

  $\begin{array}{l}x+z=-2\\ x-\frac{5}{4}=-2\\ x=-2+\frac{5}{4}\\ =\frac{-8+5}{4}=\frac{-3}{4}\end{array}$

Put value of x in eq. (ii)

  $\begin{array}{l}3(\frac{-3}{4})-2y+11(\frac{-5}{4})=-16\\ \frac{-9}{4}-2y-\frac{55}{4}=-16\\ \frac{-9}{4}-\frac{55}{4}+\frac{16}{1}=2y\\ \frac{-9-55+64}{4}=2y\\ 0=2y\\ y=0\end{array}$

Check algebraically: - Put values of $\text{x and z}$ in any equation (i)

  $\begin{array}{l}2(\frac{-3}{4})+6(\frac{-5}{4})=-9\\ \frac{-6}{4}-\frac{-30}{4}=-9\\ \frac{-36}{4}=-9\\ -9=-9\end{array}$