Chapter 7, Problem 44CR

To determine

To solve: System of linear equations.

Answer to Problem 44CR

  $x=\frac{7}{2},y=\frac{-7}{10},z=\frac{1}{10}$

Explanation of Solution

Given Information: - Given set of equations

  $x-2y+z=5,\text{ 2}x+3y+z=5,x+y+2z=3$

- The set of equations

  $x-2y+z=5(i)$

  $2x+3y+z=5(ii)$

  $x+y+2z=3(iii)$

Subtract eq (i) from eq(ii)

  $\begin{array}{l}2x+3y+z-(x-2y+z)=5-5\\ 2x-3y+z-x+2y-z=0\end{array}$

  $x+5y=0$ ----------------- (iv)

Multiply eq. (ii) by 2 and subtract it from eq.(iii)

  $\begin{array}{l}x+y+2z-2(2x+3y+z)=3-2(5)\\ x+y+2z-4x-6y-2z=3-10\end{array}$

  $-3x-5y=-7$ ----------------- (v)

Multiply eq. (iv) by 3 and add it in eq.(v)

  $\begin{array}{l}-3x-5y+3(x+5y)=-7\\ -3x-5y+3x+15y=-7\end{array}$

  $\begin{array}{l}10y=-7\\ y=\frac{-7}{10}\end{array}$

Put value of y in eq. (iv)

  $\begin{array}{l}x+5(\frac{-7}{10})=0\\ x=\frac{7}{2}\end{array}$

Put value of x and y in eq. (iii)

  $\begin{array}{l}\frac{7}{2}-\frac{7}{10}+2z=3\\ 2z=3-\frac{7}{2}+\frac{7}{10}\\ 2z=\frac{30-35+7}{10}\\ 2z=\frac{2}{10}\\ z=\frac{1}{10}\end{array}$

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

  $\begin{array}{l}\frac{7}{2}-2(\frac{-7}{10})+\frac{1}{10}=5\\ \frac{35+14+1}{10}=5\\ \frac{50}{10}=5\\ 5=5\end{array}$