Linear Algebra Examples

$- 4 x + 2 y = - 10$ , $y = 2 x - 5$

Step 1

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

$- 4 x + 2 y = - 10, y - 2 x = - 5$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} - 4 & 2 & - 10 \\ - 2 & 1 & - 5 \end{array}]$

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 $1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 2 & - \frac{1}{4} \cdot - 10 \\ - 2 & 1 & - 5 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} \\ - 2 & 1 & - 5 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} \\ - 2 + 2 \cdot 1 & 1 + 2 (- \frac{1}{2}) & - 5 + 2 (\frac{5}{2}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & 0 \end{array}]$

Step 4

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

$x - \frac{y}{2} = \frac{5}{2}$

Step 5

Solve the equation for $y$ .

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Step 5.1

Subtract $x$ from both sides of the equation.

$- \frac{y}{2} = \frac{5}{2} - x$

Step 5.2

Multiply both sides of the equation by $- 2$ .

$- 2 (- \frac{y}{2}) = - 2 (\frac{5}{2} - x)$

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

Simplify $- 2 (- \frac{y}{2})$ .

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Step 5.3.1.1.1

Cancel the common factor of $2$ .

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Step 5.3.1.1.1.1

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

$- 2 \frac{- y}{2} = - 2 (\frac{5}{2} - x)$

Step 5.3.1.1.1.2

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{- y}{2} = - 2 (\frac{5}{2} - x)$

Step 5.3.1.1.1.3

Cancel the common factor.

$2 \cdot - 1 \frac{- y}{2} = - 2 (\frac{5}{2} - x)$

Step 5.3.1.1.1.4

Rewrite the expression.

$- - y = - 2 (\frac{5}{2} - x)$

Step 5.3.1.1.2

Multiply.

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Step 5.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 y = - 2 (\frac{5}{2} - x)$

Step 5.3.1.1.2.2

Multiply $y$ by $1$ .

$y = - 2 (\frac{5}{2} - x)$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $- 2 (\frac{5}{2} - x)$ .

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Step 5.3.2.1.1

Apply the distributive property.

$y = - 2 (\frac{5}{2}) - 2 (- x)$

Step 5.3.2.1.2

Cancel the common factor of $2$ .

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Step 5.3.2.1.2.1

Factor $2$ out of $- 2$ .

$y = 2 (- 1) \frac{5}{2} - 2 (- x)$

Step 5.3.2.1.2.2

Cancel the common factor.

$y = 2 \cdot - 1 \frac{5}{2} - 2 (- x)$

Step 5.3.2.1.2.3

Rewrite the expression.

$y = - 1 \cdot 5 - 2 (- x)$

Step 5.3.2.1.3

Multiply.

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Step 5.3.2.1.3.1

Multiply $- 1$ by $5$ .

$y = - 5 - 2 (- x)$

Step 5.3.2.1.3.2

Multiply $- 1$ by $- 2$ .

$y = - 5 + 2 x$

Step 5.4

Reorder $- 5$ and $2 x$ .

$y = 2 x - 5$

Step 6

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

$(x, 2 x - 5)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ 2 x - 5 \end{array}]$