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Chapter 5: Problem 12

Solve each system. $$\left\\{\begin{array}{r} x+\quad z=3 \\ x+2 y-z=1 \\ 2 x-y+z=3 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 2\), \(y = 0\) and \(z = 1\).

Step by step solution

01

Deciding on the Variable to Eliminate First

Look at the equations, decide to eliminate \(y\) in the second and third equations because if we subtract the third equation from the second, it will be cancelled out.
02

Eliminating the Variable

Subtract the third equation from the second equation to eliminate \(y\). We have \( (x + 2y - z) - (2x - y + z) = 1 - 3 \), which simplifies to \(-x + 3y = -2\).
03

Forming a Two-Variable System

Now, we have a two-equation system with two variables, \(x\) and \(z\), from this equation and the first equation of the original system. The two equations are \(x + z = 3\) and \(-x + 3y = -2\).
04

Solving the System

Solving the two-equation system, add the two equations together to eliminate \(x\). You will find that \(z = 1\). Substitute \(z = 1\) into the first equation you get \(x + 1 = 3\), which simplifies to \(x = 2\).
05

Finding the Value of the Third Variable

Substitute \(z = 1\) and \(x = 2\) into any of the original equations to solve for \(y\). For instance, into \(x + 2y - z = 1\) we substitute to get \(2 + 2y - 1 = 1\), which simplifies to \(2y = 0\), thus \(y = 0\).

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