Linear Algebra Examples

$- x + 3 y = - 6$ , $x = 6 + 3 y$

Step 1

Subtract $3 y$ from both sides of the equation.

$- x + 3 y = - 6, x - 3 y = 6$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 3 & - 6 \\ 1 & - 3 & 6 \end{array}]$

Step 3

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 3.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 3.1.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 3 & - - 6 \\ 1 & - 3 & 6 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 3 & 6 \\ 1 & - 3 & 6 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 3.2.1

Perform the row operation $R_{2} = R_{2} - R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 3 & 6 \\ 1 - 1 & - 3 + 3 & 6 - 6 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 3 & 6 \\ 0 & 0 & 0 \end{array}]$

Step 4

Use the result matrix to declare the final solutions to the system of equations.

$x - 3 y = 6$

Step 5

Solve the equation for $y$ .

Tap for more steps...

Step 5.1

Subtract $x$ from both sides of the equation.

$- 3 y = 6 - x$

Step 5.2

Divide each term in $- 3 y = 6 - x$ by $- 3$ and simplify.

Tap for more steps...

Step 5.2.1

Divide each term in $- 3 y = 6 - x$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{6}{- 3} + \frac{- x}{- 3}$

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Cancel the common factor of $- 3$ .

Tap for more steps...

Step 5.2.2.1.1

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{6}{- 3} + \frac{- x}{- 3}$

Step 5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{6}{- 3} + \frac{- x}{- 3}$

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

Simplify each term.

Tap for more steps...

Step 5.2.3.1.1

Divide $6$ by $- 3$ .

$y = - 2 + \frac{- x}{- 3}$

Step 5.2.3.1.2

Dividing two negative values results in a positive value.

$y = - 2 + \frac{x}{3}$

Step 6

The solution is the set of ordered pairs that makes the system true.

$(x, - 2 + \frac{x}{3})$

Step 7

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ - 2 + \frac{x}{3} \end{array}]$