Question

Solve the linear equation : 2x+y-3z= 5,3 x-2y-2z= 5, and 5x-3y-z= 16.

Answer 1

Answer Question #41201, Math, Linear Algebra

$$\left\{ \begin{array}{l l} 2 x + y - 3 z = 5 & (1) \\ 3 x - 2 y - 2 z = 5 & (2) \\ 5 x - 3 y - z = 1 6 & (3) \end{array} \right.$$

We add (1) + (2):

$$\left\{ \begin{array}{l l} 2 x + y - 3 z = 5 & (1) \\ 5 x - y - 5 z = 1 0 & (2) \\ 5 x - 3 y - z = 1 6 & (3) \end{array} \right.$$

We subtract (2) from (3):

$$\left\{ \begin{array}{l l} 2 x + y - 3 z = 5 & (1) \\ 5 x - y - 5 z = 1 0 & (2) \\ - 2 y + 4 z = 6 & (3) \end{array} \right.$$

We multiply (1) by $\frac{5}{2}$ :

$$\left\{ \begin{array}{l l} 5 x + \frac {5}{2} y - \frac {1 5}{2} z = \frac {2 5}{2} & (1) \\ 5 x - y - 5 z = 1 0 & (2) \\ - 2 y + 4 z = 6 & (3) \end{array} \right.$$

We sub from (2) - (1):

$$\left\{ \begin{array}{l} 5 x + \frac {5}{2} y - \frac {1 5}{2} z = \frac {2 5}{2} \\ - \frac {7}{2} y + \frac {5}{2} z = - \frac {5}{2} \\ - 2 y + 4 z = 6 \end{array} \right. (1)$$

We multiply (2) by $\frac{4}{7}$ :

$$\left\{ \begin{array}{l} 5 x + \frac {5}{2} y - \frac {1 5}{2} z = \frac {2 5}{2} \\ - 2 y + \frac {1 0}{7} z = - \frac {1 0}{7} \\ - 2 y + 4 z = 6 \end{array} \right. (2)$$

We sub form (3) - (2):

$$\left\{ \begin{array}{l} 5 x + \frac {5}{2} y - \frac {1 5}{2} z = \frac {2 5}{2} \\ - 2 y + \frac {1 0}{7} z = - \frac {1 0}{7} \\ \frac {1 8}{7} z = \frac {5 2}{7} \end{array} \right. (3)$$

From (3), we can find $z$ :

$$\frac {1 8}{7} z = \frac {5 2}{7} \quad \Rightarrow \quad z = \frac {5 2}{7} * \frac {7}{1 8} = \frac {5 2}{1 8} = \frac {2 6}{9}$$

We can substitute $z$ to (2), and find $y$ :

$$\left\{ \begin{array}{l} 5 x + \frac {5}{2} y - \frac {1 5}{2} z = \frac {2 5}{2} \\ - 2 y + \frac {1 0}{7} * \frac {2 6}{9} = - \frac {1 0}{7} \\ z = \frac {2 6}{9} \end{array} \right. (1)$$ $$- 2 y + \frac {2 6 0}{6 3} = - \frac {1 0}{7} \quad \Rightarrow \quad - 2 y = - \frac {1 0}{7} - \frac {2 6 0}{6 3} = - \frac {3 5 0}{6 3} = - \frac {5 0}{9} \quad \Rightarrow \quad y = \frac {2 5}{9}$$

We can substitute $y$ and $z$ to (1), and find $x$:

$$\left\{ \begin{array}{l} 5x + \frac{5}{2} * \frac{25}{9} - \frac{15}{2} * \frac{26}{9} = \frac{25}{2} \quad (1) \\ \quad y = \frac{25}{9} \quad (2) \\ \quad z = \frac{26}{9} \quad (3) \end{array} \right.$$ $$5x + \frac{125}{18} - \frac{390}{18} = \frac{25}{2} \quad \Rightarrow \quad 5x - \frac{265}{18} = \frac{25}{2} \quad \Rightarrow \quad 5x = \frac{225}{18} + \frac{265}{18} = \frac{490}{18} \quad \Rightarrow \quad x = \frac{98}{18}$$

**Answer:**

$$\left\{ \begin{array}{l} x = \frac{49}{9} \\ y = \frac{25}{9} \\ z = \frac{26}{9} \end{array} \right.$$

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