Linear Algebra Examples

\frac{x + y}{7} = \frac{y + 4}{5}

$\frac{x + y}{7} = \frac{y + 4}{5}$ ,

\frac{x - z}{5} = \frac{y - 4}{2}

$\frac{x - z}{5} = \frac{y - 4}{2}$ ,

\frac{y - z}{3} = \frac{x + 2}{10}

$\frac{y - z}{3} = \frac{x + 2}{10}$

Step 1

Move all terms containing variables to the left side of the equation.

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Step 1.1

Subtract

\frac{y + 4}{5}

$\frac{y + 4}{5}$ from both sides of the equation.

\frac{x + y}{7} - \frac{y + 4}{5} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{x + y}{7} - \frac{y + 4}{5} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.2

To write

\frac{x + y}{7}

$\frac{x + y}{7}$ as a fraction with a common denominator, multiply by

\frac{5}{5}

$\frac{5}{5}$ .

\frac{x + y}{7} \cdot \frac{5}{5} - \frac{y + 4}{5} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{x + y}{7} \cdot \frac{5}{5} - \frac{y + 4}{5} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.3

To write

- \frac{y + 4}{5}

$- \frac{y + 4}{5}$ as a fraction with a common denominator, multiply by

\frac{7}{7}

$\frac{7}{7}$ .

\frac{x + y}{7} \cdot \frac{5}{5} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{x + y}{7} \cdot \frac{5}{5} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.4

Write each expression with a common denominator of

35

$35$ , by multiplying each by an appropriate factor of

1

$1$ .

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Step 1.4.1

Multiply

\frac{x + y}{7}

$\frac{x + y}{7}$ by

\frac{5}{5}

$\frac{5}{5}$ .

\frac{(x + y) \cdot 5}{7 \cdot 5} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{(x + y) \cdot 5}{7 \cdot 5} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.4.2

Multiply

7

$7$ by

5

$5$ .

\frac{(x + y) \cdot 5}{35} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{(x + y) \cdot 5}{35} - \frac{y + 4}{5} \cdot \frac{7}{7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.4.3

Multiply

\frac{y + 4}{5}

$\frac{y + 4}{5}$ by

\frac{7}{7}

$\frac{7}{7}$ .

\frac{(x + y) \cdot 5}{35} - \frac{(y + 4) \cdot 7}{5 \cdot 7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}

$\frac{(x + y) \cdot 5}{35} - \frac{(y + 4) \cdot 7}{5 \cdot 7} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.4.4

Multiply

$5$ by

$7$ .

$\frac{(x + y) \cdot 5}{35} - \frac{(y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.5

Combine the numerators over the common denominator.

$\frac{(x + y) \cdot 5 - (y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Apply the distributive property.

$\frac{x \cdot 5 + y \cdot 5 - (y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.2

Move

$5$ to the left of

$x$ .

$\frac{5 \cdot x + y \cdot 5 - (y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.3

Move

$5$ to the left of

$y$ .

$\frac{5 \cdot x + 5 \cdot y - (y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.4

Multiply

$5$ by

$y$ .

$\frac{5 x + 5 y - (y + 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.5

Apply the distributive property.

$\frac{5 x + 5 y + (- y - 1 \cdot 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.6

Multiply

$- 1$ by

$4$ .

$\frac{5 x + 5 y + (- y - 4) \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.7

Apply the distributive property.

$\frac{5 x + 5 y - y \cdot 7 - 4 \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.8

Multiply

$7$ by

$- 1$ .

$\frac{5 x + 5 y - 7 y - 4 \cdot 7}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.9

Multiply

$- 4$ by

$7$ .

$\frac{5 x + 5 y - 7 y - 28}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 1.6.10

Subtract

$7 y$ from

$5 y$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{x - z}{5} = \frac{y - 4}{2}, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2

Move all terms containing variables to the left side of the equation.

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Step 2.1

Subtract

$\frac{y - 4}{2}$ from both sides of the equation.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{x - z}{5} - \frac{y - 4}{2} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.2

To write

$\frac{x - z}{5}$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{x - z}{5} \cdot \frac{2}{2} - \frac{y - 4}{2} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.3

To write

$- \frac{y - 4}{2}$ as a fraction with a common denominator, multiply by

$\frac{5}{5}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{x - z}{5} \cdot \frac{2}{2} - \frac{y - 4}{2} \cdot \frac{5}{5} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.4

Write each expression with a common denominator of

$10$ , by multiplying each by an appropriate factor of

$1$ .

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Step 2.4.1

Multiply

$\frac{x - z}{5}$ by

$\frac{2}{2}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{(x - z) \cdot 2}{5 \cdot 2} - \frac{y - 4}{2} \cdot \frac{5}{5} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.4.2

Multiply

$5$ by

$2$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{(x - z) \cdot 2}{10} - \frac{y - 4}{2} \cdot \frac{5}{5} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.4.3

Multiply

$\frac{y - 4}{2}$ by

$\frac{5}{5}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{(x - z) \cdot 2}{10} - \frac{(y - 4) \cdot 5}{2 \cdot 5} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.4.4

Multiply

$2$ by

$5$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{(x - z) \cdot 2}{10} - \frac{(y - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.5

Combine the numerators over the common denominator.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{(x - z) \cdot 2 - (y - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{x \cdot 2 - z \cdot 2 - (y - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.2

Move

$2$ to the left of

$x$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 \cdot x - z \cdot 2 - (y - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.3

Multiply

$2$ by

$- 1$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 \cdot x - 2 z - (y - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.4

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z + (- y - - 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.5

Multiply

$- 1$ by

$- 4$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z + (- y + 4) \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.6

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - y \cdot 5 + 4 \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.7

Multiply

$5$ by

$- 1$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 4 \cdot 5}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 2.6.8

Multiply

$4$ by

$5$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{y - z}{3} = \frac{x + 2}{10}$

Step 3

Move all terms containing variables to the left side of the equation.

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Step 3.1

Subtract

$\frac{x + 2}{10}$ from both sides of the equation.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{y - z}{3} - \frac{x + 2}{10} = 0$

Step 3.2

To write

$\frac{y - z}{3}$ as a fraction with a common denominator, multiply by

$\frac{10}{10}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{y - z}{3} \cdot \frac{10}{10} - \frac{x + 2}{10} = 0$

Step 3.3

To write

$- \frac{x + 2}{10}$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{y - z}{3} \cdot \frac{10}{10} - \frac{x + 2}{10} \cdot \frac{3}{3} = 0$

Step 3.4

Write each expression with a common denominator of

$30$ , by multiplying each by an appropriate factor of

$1$ .

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Step 3.4.1

Multiply

$\frac{y - z}{3}$ by

$\frac{10}{10}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{(y - z) \cdot 10}{3 \cdot 10} - \frac{x + 2}{10} \cdot \frac{3}{3} = 0$

Step 3.4.2

Multiply

$3$ by

$10$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{(y - z) \cdot 10}{30} - \frac{x + 2}{10} \cdot \frac{3}{3} = 0$

Step 3.4.3

Multiply

$\frac{x + 2}{10}$ by

$\frac{3}{3}$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{(y - z) \cdot 10}{30} - \frac{(x + 2) \cdot 3}{10 \cdot 3} = 0$

Step 3.4.4

Multiply

$10$ by

$3$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{(y - z) \cdot 10}{30} - \frac{(x + 2) \cdot 3}{30} = 0$

Step 3.5

Combine the numerators over the common denominator.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{(y - z) \cdot 10 - (x + 2) \cdot 3}{30} = 0$

Step 3.6

Simplify the numerator.

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Step 3.6.1

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{y \cdot 10 - z \cdot 10 - (x + 2) \cdot 3}{30} = 0$

Step 3.6.2

Move

$10$ to the left of

$y$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 \cdot y - z \cdot 10 - (x + 2) \cdot 3}{30} = 0$

Step 3.6.3

Multiply

$10$ by

$- 1$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 \cdot y - 10 z - (x + 2) \cdot 3}{30} = 0$

Step 3.6.4

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 y - 10 z + (- x - 1 \cdot 2) \cdot 3}{30} = 0$

Step 3.6.5

Multiply

$- 1$ by

$2$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 y - 10 z + (- x - 2) \cdot 3}{30} = 0$

Step 3.6.6

Apply the distributive property.

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 y - 10 z - x \cdot 3 - 2 \cdot 3}{30} = 0$

Step 3.6.7

Multiply

$3$ by

$- 1$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 y - 10 z - 3 x - 2 \cdot 3}{30} = 0$

Step 3.6.8

Multiply

$- 2$ by

$3$ .

$\frac{5 x - 2 y - 28}{35} = 0, \frac{2 x - 2 z - 5 y + 20}{10} = 0, \frac{10 y - 10 z - 3 x - 6}{30} = 0$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{35} & 0 & 0 & 0 \\ \frac{1}{10} & 0 & 0 & 0 \\ 0 & \frac{1}{30} & 0 & 0 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Multiply each element of

$R_{1}$ by

$35$ to make the entry at

$1, 1$ a

$1$ .

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Step 5.1.1

Multiply each element of

$R_{1}$ by

$35$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} 35 (\frac{1}{35}) & 35 \cdot 0 & 35 \cdot 0 & 35 \cdot 0 \\ \frac{1}{10} & 0 & 0 & 0 \\ 0 & \frac{1}{30} & 0 & 0 \end{array}]$

Step 5.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ \frac{1}{10} & 0 & 0 & 0 \\ 0 & \frac{1}{30} & 0 & 0 \end{array}]$

Step 5.2

Perform the row operation

$R_{2} = R_{2} - \frac{1}{10} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 5.2.1

Perform the row operation

$R_{2} = R_{2} - \frac{1}{10} R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ \frac{1}{10} - \frac{1}{10} \cdot 1 & 0 - \frac{1}{10} \cdot 0 & 0 - \frac{1}{10} \cdot 0 & 0 - \frac{1}{10} \cdot 0 \\ 0 & \frac{1}{30} & 0 & 0 \end{array}]$

Step 5.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & \frac{1}{30} & 0 & 0 \end{array}]$

Step 5.3

Swap

$R_{3}$ with

$R_{2}$ to put a nonzero entry at

$2, 2$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{30} & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.4

Multiply each element of

$R_{2}$ by

$30$ to make the entry at

$2, 2$ a

$1$ .

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Step 5.4.1

Multiply each element of

$R_{2}$ by

$30$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 30 \cdot 0 & 30 (\frac{1}{30}) & 30 \cdot 0 & 30 \cdot 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.4.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 6

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

$x = 0$

$y = 0$

Step 7

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

$(0, 0, z)$

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.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 0 \\ 0 \\ z \end{array}]$