Linear Algebra Examples

$2 y = 14 - 6 x$ , $3 x + y = 7$

Step 1

Add $6 x$ to both sides of the equation.

$2 y + 6 x = 14, 3 x + y = 7$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 6 & 2 & 14 \\ 3 & 1 & 7 \end{array}]$

Step 3

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{2}{6} & \frac{14}{6} \\ 3 & 1 & 7 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{7}{3} \\ 3 & 1 & 7 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{7}{3} \\ 3 - 3 \cdot 1 & 1 - 3 (\frac{1}{3}) & 7 - 3 (\frac{7}{3}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{7}{3} \\ 0 & 0 & 0 \end{array}]$

Step 4

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

$x + \frac{y}{3} = \frac{7}{3}$

Step 5

Solve the equation for $y$ .

Tap for more steps...

Step 5.1

Subtract $x$ from both sides of the equation.

$\frac{y}{3} = \frac{7}{3} - x$

Step 5.2

Multiply both sides of the equation by $3$ .

$3 \frac{y}{3} = 3 (\frac{7}{3} - x)$

Step 5.3

Simplify both sides of the equation.

Tap for more steps...

Step 5.3.1

Simplify the left side.

Tap for more steps...

Step 5.3.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.1.1.1

Cancel the common factor.

$3 \frac{y}{3} = 3 (\frac{7}{3} - x)$

Step 5.3.1.1.2

Rewrite the expression.

$y = 3 (\frac{7}{3} - x)$

Step 5.3.2

Simplify the right side.

Tap for more steps...

Step 5.3.2.1

Simplify $3 (\frac{7}{3} - x)$ .

Tap for more steps...

Step 5.3.2.1.1

Apply the distributive property.

$y = 3 (\frac{7}{3}) + 3 (- x)$

Step 5.3.2.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.2.1.2.1

Cancel the common factor.

$y = 3 (\frac{7}{3}) + 3 (- x)$

Step 5.3.2.1.2.2

Rewrite the expression.

$y = 7 + 3 (- x)$

Step 5.3.2.1.3

Multiply $- 1$ by $3$ .

$y = 7 - 3 x$

Step 5.4

Reorder $7$ and $- 3 x$ .

$y = - 3 x + 7$

Step 6

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

$(x, - 3 x + 7)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ - 3 x + 7 \end{array}]$