Linear Algebra Examples

$y = \frac{- 2}{3} x + 4$ , $2 x + 3 y = 12$

Step 1

Simplify each term.

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Step 1.1

Move the negative in front of the fraction.

$y = - \frac{2}{3} x + 4, 2 x + 3 y = 12$

Step 1.2

Combine $x$ and $\frac{2}{3}$ .

$y = - \frac{x \cdot 2}{3} + 4, 2 x + 3 y = 12$

Step 1.3

Move $2$ to the left of $x$ .

$y = - \frac{2 x}{3} + 4, 2 x + 3 y = 12$

Step 2

Add $\frac{2 x}{3}$ to both sides of the equation.

$y + \frac{2 x}{3} = 4, 2 x + 3 y = 12$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{2}{3} & 1 & 4 \\ 2 & 3 & 12 \end{array}]$

Step 4

Row reduce to eliminate one of the variables.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{3}{2}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{3}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{3}{2} \cdot \frac{2}{3} & \frac{3}{2} \cdot 1 & \frac{3}{2} \cdot 4 \\ 2 & 3 & 12 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 6 \\ 2 & 3 & 12 \end{array}]$

Step 4.2

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 4.2.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 6 \\ 2 - 2 \cdot 1 & 3 - 2 (\frac{3}{2}) & 12 - 2 \cdot 6 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 6 \\ 0 & 0 & 0 \end{array}]$

Step 5

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

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

Step 6

Solve the equation for $y$ .

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Step 6.1

Subtract $x$ from both sides of the equation.

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

Step 6.2

Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 y}{2} = \frac{2}{3} (6 - x)$

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 $\frac{2}{3} \cdot \frac{3 y}{2}$ .

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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

Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 y}{2} = \frac{2}{3} (6 - x)$

Step 6.3.1.1.1.2

Rewrite the expression.

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

Step 6.3.1.1.2

Cancel the common factor of $3$ .

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Step 6.3.1.1.2.1

Factor $3$ out of $3 y$ .

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

Step 6.3.1.1.2.2

Cancel the common factor.

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

Step 6.3.1.1.2.3

Rewrite the expression.

$y = \frac{2}{3} (6 - x)$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Simplify $\frac{2}{3} (6 - x)$ .

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Step 6.3.2.1.1

Apply the distributive property.

$y = \frac{2}{3} \cdot 6 + \frac{2}{3} (- x)$

Step 6.3.2.1.2

Cancel the common factor of $3$ .

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Step 6.3.2.1.2.1

Factor $3$ out of $6$ .

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

Step 6.3.2.1.2.2

Cancel the common factor.

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

Step 6.3.2.1.2.3

Rewrite the expression.

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

Step 6.3.2.1.3

Multiply $2$ by $2$ .

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

Step 6.3.2.1.4

Combine $\frac{2}{3}$ and $x$ .

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

Step 6.4

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

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

Step 7

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

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

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.

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