Linear Algebra Examples

$2 x - y = 10$ , $x = \frac{1}{2} y + 5$

Step 1

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

$2 x - y = 10, x = \frac{y}{2} + 5$

Step 2

Subtract $\frac{y}{2}$ from both sides of the equation.

$2 x - y = 10, x - \frac{y}{2} = 5$

Step 3

Write the system of equations in matrix form.

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

Step 4

Row reduce to eliminate one of the variables.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 5

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

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

Step 6

Solve the equation for $y$ .

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Step 6.1

Subtract $x$ from both sides of the equation.

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

Step 6.2

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

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

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

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

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Step 6.3.1.1.1

Cancel the common factor of $2$ .

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Step 6.3.1.1.1.1

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

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

Step 6.3.1.1.1.2

Factor $2$ out of $- 2$ .

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

Step 6.3.1.1.1.3

Cancel the common factor.

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

Step 6.3.1.1.1.4

Rewrite the expression.

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

Step 6.3.1.1.2

Multiply.

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Step 6.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

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

Step 6.3.1.1.2.2

Multiply $y$ by $1$ .

$y = - 2 (5 - x)$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Simplify $- 2 (5 - x)$ .

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Step 6.3.2.1.1

Apply the distributive property.

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

Step 6.3.2.1.2

Multiply.

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Step 6.3.2.1.2.1

Multiply $- 2$ by $5$ .

$y = - 10 - 2 (- x)$

Step 6.3.2.1.2.2

Multiply $- 1$ by $- 2$ .

$y = - 10 + 2 x$

Step 6.4

Reorder $- 10$ and $2 x$ .

$y = 2 x - 10$

Step 7

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

$(x, 2 x - 10)$

Step 8

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ 2 x - 10 \end{array}]$