Solve each sys tem u sing Eliminat 1) x-y+z=3 2x+y+z=8 3x+y-z=1

To solve the system of equations using the elimination method, we aim to eliminate one variable at a time. Let's work through the given system:<br /><br />\[<br />\begin{array}{rcl}<br />1) & x - y + z &= 3 \\<br />2) & 2x + y + z &= 8 \\<br />3) & 3x + y - z &= 1<br />\end{array}<br />\]<br /><br />**Step 1: Eliminate \(z\) from Equations (1) and (2).**<br /><br />Add Equation (1) and Equation (2):<br /><br />\[<br />(x - y + z) + (2x + y + z) = 3 + 8<br />\]<br /><br />This simplifies to:<br /><br />\[<br />3x + 2z = 11 \quad \text{(Equation 4)}<br />\]<br /><br />**Step 2: Eliminate \(z\) from Equations (1) and (3).**<br /><br />Add Equation (1) and Equation (3):<br /><br />\[<br />(x - y + z) + (3x + y - z) = 3 + 1<br />\]<br /><br />This simplifies to:<br /><br />\[<br />4x = 4 \quad \Rightarrow \quad x = 1 \quad \text{(Equation 5)}<br />\]<br /><br />**Step 3: Substitute \(x = 1\) into Equation (4).**<br /><br />Substitute \(x = 1\) into Equation (4):<br /><br />\[<br />3(1) + 2z = 11<br />\]<br /><br />This simplifies to:<br /><br />\[<br />3 + 2z = 11 \quad \Rightarrow \quad 2z = 8 \quad \Rightarrow \quad z = 4<br />\]<br /><br />**Step 4: Substitute \(x = 1\) and \(z = 4\) back into Equation (1) to find \(y\).**<br /><br />Substitute \(x = 1\) and \(z = 4\) into Equation (1):<br /><br />\[<br />1 - y + 4 = 3<br />\]<br /><br />This simplifies to:<br /><br />\[<br />5 - y = 3 \quad \Rightarrow \quad -y = -2 \quad \Rightarrow \quad y = 2<br />\]<br /><br />Thus, the solution to the system is:<br /><br />\[<br />x = 1, \quad y = 2, \quad z = 4<br />\]