Linear Algebra Examples

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

Step 1

Add $2 x$ to both sides of the equation.

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

Step 2

Write the system of equations in matrix form.

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

Step 3

Row reduce to eliminate one of the variables.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{2}$ .

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

Step 4

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

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

Step 5

Solve the equation for $y$ .

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Step 5.1

Subtract $x$ from both sides of the equation.

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

Step 5.2

Multiply both sides of the equation by $2$ .

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

Step 5.3

Simplify both sides of the equation.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Cancel the common factor of $2$ .

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Step 5.3.1.1.1

Cancel the common factor.

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

Step 5.3.1.1.2

Rewrite the expression.

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

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Apply the distributive property.

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

Step 5.3.2.1.2

Cancel the common factor of $2$ .

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Step 5.3.2.1.2.1

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

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

Step 5.3.2.1.2.2

Cancel the common factor.

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

Step 5.3.2.1.2.3

Rewrite the expression.

$y = - 3 + 2 (- x)$

Step 5.3.2.1.3

Multiply $- 1$ by $2$ .

$y = - 3 - 2 x$

Step 5.4

Reorder $- 3$ and $- 2 x$ .

$y = - 2 x - 3$

Step 6

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

$(x, - 2 x - 3)$

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.

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