) -x_(1)-x_(2)-2x_(4)+2x_(5)=1 2x_(1)-x_(2)+x_(3)-2x_(4)+2x_(5)=-1 x_(1)-2x_(2)-x_(4)-x_(3)=0 2x_(1)

To solve the given system of linear equations, we can use the method of substitution or elimination. Let's solve it using the elimination method.<br /><br />The given system of equations is:<br />\[<br />\begin{align*}<br />-x_{1}-x_{2}-2x_{4}+2x_{5} &= 1 \quad \text{(1)} \\<br />2x_{1}-x_{2}+x_{3}-2x_{4}+2x_{5} &= -1 \quad \text{(2)} \\<br />x_{1}-2x_{2}-x_{4}-x_{3} &= 0 \quad \text{(3)} \\<br />2x_{1}+x_{2}+x_{3}-2x_{4}+2x_{5} &= 1 \quad \text{(4)}<br />\end{align*}<br />\]<br /><br />Step 1: Add equation (1) and equation (4) to eliminate \(x_{2}\):<br />\[<br />\begin{align*}<br />(-x_{1}-x_{2}-2x_{4}+2x_{5}) + (2x_{1}+x_{2}+x_{3}-2x_{4}+2x_{5}) &= 1 + 1 \\<br />x_{1}+x_{3}+4x_{5} &= 2 \quad \text{(5)}<br />\end{align*}<br />\]<br /><br />Step 2: Multiply equation (3) by 2 and add it to equation (2) to eliminate \(x_{3}\):<br />\[<br />\begin{align*}<br />2(x_{1}-2x_{2}-x_{4}-x_{3}) + (2x_{1}-x_{2}+x_{3}-2x_{4}+2x_{5}) &= 2(0) + (-1) \\<br />4x_{1}-5x_{2}-4x_{4}+2x_{5} &= -1 \quad \text{(6)}<br />\end{align*}<br />\]<br /><br />Step 3: Multiply equation (1) by 2 and add it to equation (6) to eliminate \(x_{1}\):<br />\[<br />\begin{align*}<br />2(-x_{1}-x_{2}-2x_{4}+2x_{5}) + (4x_{1}-5x_{2}-4x_{4}+2x_{5}) &= 2(1) + (-1) \\<br />-2x_{2} &= 1 \\<br />x_{2} &= -\frac{1}{2} \quad \text{(7)}<br />\end{align*}<br />\]<br /><br />Step 4: Substitute \(x_{2}\) into equation (5) to find \(x_{1}\) and \(x_{3}\):<br />\[<br />\begin{align*}<br />x_{1}+x_{3}+4x_{5} &= 2 \\<br />x_{1} &= 2 - x_{3} - 4x_{5} \quad \text{(8)}<br />\end{align*}<br />\]<br /><br />Step 5: Substitute \(x_{2}\) into equation (6) to find \(x_{4}\) and \(x_{5}\):<br />\[<br />\begin{align*}<br />4x_{1}-5x_{2}-4x_{4}+2x_{5} &= -1 \\<br />4(2 - x_{3} - 4x_{5}) - 5(-\frac{1}{2}) - 4x_{4} + 2x_{5} &= -1 \\<br />8 - 4x_{3} - 16x_{5} + \frac{5}{2} - 4x_{4} + 2x_{5} &= -1 \\<br />-4x_{3} - 14x_{5} - 4x_{4} &= -\frac{21}{2} \\<br />x_{3} &= \frac{1}{4} - \frac{7}{2}x_{5} + x_{4} \quad \text{(9)}<br />\end{align*}<br />\]<br /><br />Step 6: Substitute \(x_{2}\) and \(x_{3}\) into equation (8) to find \(x_{1}\):<br />\[<br />\begin{align*}<br />x_{1} &= 2 - (\frac{1}{4} - \frac{7}{2}x_{5} + x_{4}) - 4x_{5} \\<br />x_{1} &= \frac{7}{4} - \frac{15}{2}x_{5} - x_{4} \quad \text{(10)}<br />\end{align*}<br />\]<br /><br />Therefore, the solution to the given system of equations is:<br />\[<br />\begin{align*}<br />x_{1} &= \frac