Linear Algebra Examples

$16 x - 10 y = 6$ , $8 x - 5 y = 3$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 16 & - 10 & 6 \\ 8 & - 5 & 3 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{16}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{16}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{16}{16} & \frac{- 10}{16} & \frac{6}{16} \\ 8 & - 5 & 3 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{3}{8} \\ 8 & - 5 & 3 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{3}{8} \\ 8 - 8 \cdot 1 & - 5 - 8 (- \frac{5}{8}) & 3 - 8 (\frac{3}{8}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{8} & \frac{3}{8} \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x - \frac{5 y}{8} = \frac{3}{8}$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$- \frac{5 y}{8} = \frac{3}{8} - x$

Step 4.2

Multiply both sides of the equation by $- \frac{8}{5}$ .

$- \frac{8}{5} (- \frac{5 y}{8}) = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3

Simplify both sides of the equation.

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Step 4.3.1

Simplify the left side.

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Step 4.3.1.1

Simplify $- \frac{8}{5} (- \frac{5 y}{8})$ .

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Step 4.3.1.1.1

Cancel the common factor of $8$ .

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Step 4.3.1.1.1.1

Move the leading negative in $- \frac{8}{5}$ into the numerator.

$\frac{- 8}{5} (- \frac{5 y}{8}) = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.1.2

Move the leading negative in $- \frac{5 y}{8}$ into the numerator.

$\frac{- 8}{5} \cdot \frac{- 5 y}{8} = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.1.3

Factor $8$ out of $- 8$ .

$\frac{8 (- 1)}{5} \cdot \frac{- 5 y}{8} = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.1.4

Cancel the common factor.

$\frac{8 \cdot - 1}{5} \cdot \frac{- 5 y}{8} = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{5} (- 5 y) = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.2

Cancel the common factor of $5$ .

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Step 4.3.1.1.2.1

Factor $5$ out of $- 5 y$ .

$\frac{- 1}{5} (5 (- y)) = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{5} (5 (- y)) = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.2.3

Rewrite the expression.

$- - y = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.3

Multiply.

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Step 4.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.1.1.3.2

Multiply $y$ by $1$ .

$y = - \frac{8}{5} (\frac{3}{8} - x)$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Simplify $- \frac{8}{5} (\frac{3}{8} - x)$ .

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Step 4.3.2.1.1

Apply the distributive property.

$y = - \frac{8}{5} \cdot \frac{3}{8} - \frac{8}{5} (- x)$

Step 4.3.2.1.2

Cancel the common factor of $8$ .

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Step 4.3.2.1.2.1

Move the leading negative in $- \frac{8}{5}$ into the numerator.

$y = \frac{- 8}{5} \cdot \frac{3}{8} - \frac{8}{5} (- x)$

Step 4.3.2.1.2.2

Factor $8$ out of $- 8$ .

$y = \frac{8 (- 1)}{5} \cdot \frac{3}{8} - \frac{8}{5} (- x)$

Step 4.3.2.1.2.3

Cancel the common factor.

$y = \frac{8 \cdot - 1}{5} \cdot \frac{3}{8} - \frac{8}{5} (- x)$

Step 4.3.2.1.2.4

Rewrite the expression.

$y = \frac{- 1}{5} \cdot 3 - \frac{8}{5} (- x)$

Step 4.3.2.1.3

Combine $\frac{- 1}{5}$ and $3$ .

$y = \frac{- 1 \cdot 3}{5} - \frac{8}{5} (- x)$

Step 4.3.2.1.4

Multiply $- 1$ by $3$ .

$y = \frac{- 3}{5} - \frac{8}{5} (- x)$

Step 4.3.2.1.5

Multiply $- \frac{8}{5} (- x)$ .

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Step 4.3.2.1.5.1

Multiply $- 1$ by $- 1$ .

$y = \frac{- 3}{5} + 1 (\frac{8}{5}) x$

Step 4.3.2.1.5.2

Multiply $\frac{8}{5}$ by $1$ .

$y = \frac{- 3}{5} + \frac{8}{5} x$

Step 4.3.2.1.5.3

Combine $\frac{8}{5}$ and $x$ .

$y = \frac{- 3}{5} + \frac{8 x}{5}$

Step 4.3.2.1.6

Move the negative in front of the fraction.

$y = - \frac{3}{5} + \frac{8 x}{5}$

Step 4.4

Reorder $- \frac{3}{5}$ and $\frac{8 x}{5}$ .

$y = \frac{8 x}{5} - \frac{3}{5}$

Step 5

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

$(x, \frac{8 x}{5} - \frac{3}{5})$

Step 6

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ \frac{8 x}{5} - \frac{3}{5} \end{array}]$