Chapter 6.1, Problem 78SR

To determine

To solve: The given system of equations.

Answer to Problem 78SR

The solution for the given system of equations is

  $\begin{array}{l}a=4\\ b=-2\\ c=1\end{array}$

Explanation of Solution

Given:

The given system of equations is

  $\begin{array}{l}2a+b-c=5\mathrm{...}\left(1\right)\\ a-b+3c=9\mathrm{...}\left(2\right)\\ 3a-6c=6\mathrm{...}\left(3\right)\end{array}$

Calculation:

Separate a from the third equation.

  $\begin{array}{l}3a=6+6c\\ a=2\left(c+1\right)\end{array}$

Substitute this value of a in the first equation.

  $\begin{array}{l}2\left(2\left(c+1\right)\right)+b-c=5\\ 2\left(c+1\right)-b+3c=9\end{array}$

Simplifying the above two equations resultsin

  $\begin{array}{l}3c+b+4=5\\ 5c-b+2=9\end{array}$

Separate c from the above equation $3c+b+4=5$ and substitute in

  $5c-b+2=9$ .

  $\begin{array}{l}c=\frac{-b+1}{3}\\ \text{Substitute the value of c in equation }5c-b+2=9\\ 5\times \frac{-b+1}{3}-b+2=9\\ \frac{-8b+5}{3}+2=9\\ \frac{-8b+5}{3}=7\\ -8b+5=21\\ -8b=16\\ b=-2\end{array}$

Calculate the value of c .

  $\begin{array}{c}c=\frac{-b+1}{3}\\ =\frac{-\left(-2\right)+1}{3}\\ =1\end{array}$

Calculate the value of a .

  $\begin{array}{c}a=2\left(c+1\right)\\ =2\left(1+1\right)\\ =4\end{array}$

Therefore, the solution for the given system of equation is

  $\begin{array}{l}a=4\\ b=-2\\ c=1\end{array}$