Linear Algebra Examples

$- 3 x - 4 y = 2$ , $8 y = - 6 x - 4$

Step 1

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

$- 3 x - 4 y = 2, 8 y + 6 x = - 4$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} - 3 & - 4 & 2 \\ 6 & 8 & - 4 \end{array}]$

Step 3

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 4 & - \frac{1}{3} \cdot 2 \\ 6 & 8 & - 4 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ 6 & 8 & - 4 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ 6 - 6 \cdot 1 & 8 - 6 (\frac{4}{3}) & - 4 - 6 (- \frac{2}{3}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & - \frac{2}{3} \\ 0 & 0 & 0 \end{array}]$

Step 4

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

$x + \frac{4 y}{3} = - \frac{2}{3}$

Step 5

Solve the equation for $y$ .

Tap for more steps...

Step 5.1

Subtract $x$ from both sides of the equation.

$\frac{4 y}{3} = - \frac{2}{3} - x$

Step 5.2

Multiply both sides of the equation by $\frac{3}{4}$ .

$\frac{3}{4} \cdot \frac{4 y}{3} = \frac{3}{4} (- \frac{2}{3} - x)$

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

Simplify $\frac{3}{4} \cdot \frac{4 y}{3}$ .

Tap for more steps...

Step 5.3.1.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.1.1.1.1

Cancel the common factor.

$\frac{3}{4} \cdot \frac{4 y}{3} = \frac{3}{4} (- \frac{2}{3} - x)$

Step 5.3.1.1.1.2

Rewrite the expression.

$\frac{1}{4} (4 y) = \frac{3}{4} (- \frac{2}{3} - x)$

Step 5.3.1.1.2

Cancel the common factor of $4$ .

Tap for more steps...

Step 5.3.1.1.2.1

Factor $4$ out of $4 y$ .

$\frac{1}{4} (4 (y)) = \frac{3}{4} (- \frac{2}{3} - x)$

Step 5.3.1.1.2.2

Cancel the common factor.

$\frac{1}{4} (4 y) = \frac{3}{4} (- \frac{2}{3} - x)$

Step 5.3.1.1.2.3

Rewrite the expression.

$y = \frac{3}{4} (- \frac{2}{3} - x)$

Step 5.3.2

Simplify the right side.

Tap for more steps...

Step 5.3.2.1

Simplify $\frac{3}{4} (- \frac{2}{3} - x)$ .

Tap for more steps...

Step 5.3.2.1.1

Apply the distributive property.

$y = \frac{3}{4} (- \frac{2}{3}) + \frac{3}{4} (- x)$

Step 5.3.2.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.3.2.1.2.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$y = \frac{3}{4} \cdot \frac{- 2}{3} + \frac{3}{4} (- x)$

Step 5.3.2.1.2.2

Cancel the common factor.

$y = \frac{3}{4} \cdot \frac{- 2}{3} + \frac{3}{4} (- x)$

Step 5.3.2.1.2.3

Rewrite the expression.

$y = \frac{1}{4} \cdot - 2 + \frac{3}{4} (- x)$

Step 5.3.2.1.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.3.2.1.3.1

Factor $2$ out of $4$ .

$y = \frac{1}{2 (2)} \cdot - 2 + \frac{3}{4} (- x)$

Step 5.3.2.1.3.2

Factor $2$ out of $- 2$ .

$y = \frac{1}{2 \cdot 2} \cdot (2 \cdot - 1) + \frac{3}{4} (- x)$

Step 5.3.2.1.3.3

Cancel the common factor.

$y = \frac{1}{2 \cdot 2} \cdot (2 \cdot - 1) + \frac{3}{4} (- x)$

Step 5.3.2.1.3.4

Rewrite the expression.

$y = \frac{1}{2} \cdot - 1 + \frac{3}{4} (- x)$

Step 5.3.2.1.4

Combine $\frac{1}{2}$ and $- 1$ .

$y = \frac{- 1}{2} + \frac{3}{4} (- x)$

Step 5.3.2.1.5

Combine $\frac{3}{4}$ and $x$ .

$y = \frac{- 1}{2} - \frac{3 x}{4}$

Step 5.3.2.1.6

Move the negative in front of the fraction.

$y = - \frac{1}{2} - \frac{3 x}{4}$

Step 5.4

Reorder $- \frac{1}{2}$ and $- \frac{3 x}{4}$ .

$y = - \frac{3 x}{4} - \frac{1}{2}$

Step 6

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

$(x, - \frac{3 x}{4} - \frac{1}{2})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ - \frac{3 x}{4} - \frac{1}{2} \end{array}]$