Linear Algebra Examples

$(4 x - y) = 6$ , $(\frac{1}{2} y) = - 3 + 2 x$

Step 1

Remove parentheses.

$4 x - y = 6, \frac{1}{2} y = - 3 + 2 x$

Step 2

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

$4 x - y = 6, \frac{y}{2} = - 3 + 2 x$

Step 3

Subtract $2 x$ from both sides of the equation.

$4 x - y = 6, \frac{y}{2} - 2 x = - 3$

Step 4

Write the system of equations in matrix form.

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

Step 5

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 5.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 5.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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

Step 5.1.2

Simplify $R_{1}$ .

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

Step 5.2

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

Tap for more steps...

Step 5.2.1

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

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

Step 5.2.2

Simplify $R_{2}$ .

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

Step 6

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

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

Step 7

Solve the equation for $y$ .

Tap for more steps...

Step 7.1

Subtract $x$ from both sides of the equation.

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

Step 7.2

Multiply both sides of the equation by $- 4$ .

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

Step 7.3

Simplify both sides of the equation.

Tap for more steps...

Step 7.3.1

Simplify the left side.

Tap for more steps...

Step 7.3.1.1

Simplify $- 4 (- \frac{y}{4})$ .

Tap for more steps...

Step 7.3.1.1.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 7.3.1.1.1.1

Move the leading negative in $- \frac{y}{4}$ into the numerator.

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

Step 7.3.1.1.1.2

Factor $4$ out of $- 4$ .

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

Step 7.3.1.1.1.3

Cancel the common factor.

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

Step 7.3.1.1.1.4

Rewrite the expression.

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

Step 7.3.1.1.2

Multiply.

Tap for more steps...

Step 7.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 7.3.1.1.2.2

Multiply $y$ by $1$ .

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

Step 7.3.2

Simplify the right side.

Tap for more steps...

Step 7.3.2.1

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

Tap for more steps...

Step 7.3.2.1.1

Apply the distributive property.

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

Step 7.3.2.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 7.3.2.1.2.1

Factor $2$ out of $- 4$ .

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

Step 7.3.2.1.2.2

Cancel the common factor.

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

Step 7.3.2.1.2.3

Rewrite the expression.

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

Step 7.3.2.1.3

Multiply.

Tap for more steps...

Step 7.3.2.1.3.1

Multiply $- 2$ by $3$ .

$y = - 6 - 4 (- x)$

Step 7.3.2.1.3.2

Multiply $- 1$ by $- 4$ .

$y = - 6 + 4 x$

Step 7.4

Reorder $- 6$ and $4 x$ .

$y = 4 x - 6$

Step 8

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

$(x, 4 x - 6)$

Step 9

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ 4 x - 6 \end{array}]$