Linear Algebra Examples

$\frac{2 x + 5}{17} - (5 - y) = 60$ , $\frac{y + 62}{2} - (1 - x) = 40$

Step 1

Simplify.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$\frac{2 x + 5}{17} - 1 \cdot 5 - - y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.1.2

Multiply $- 1$ by $5$ .

$\frac{2 x + 5}{17} - 5 - - y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.1.3

Multiply $- - y$ .

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Step 1.1.3.1

Multiply $- 1$ by $- 1$ .

$\frac{2 x + 5}{17} - 5 + 1 y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.1.3.2

Multiply $y$ by $1$ .

$\frac{2 x + 5}{17} - 5 + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.2

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{17}{17}$ .

$\frac{2 x + 5}{17} - 5 \cdot \frac{17}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.3

Combine $- 5$ and $\frac{17}{17}$ .

$\frac{2 x + 5}{17} + \frac{- 5 \cdot 17}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.4

Combine the numerators over the common denominator.

$\frac{2 x + 5 - 5 \cdot 17}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.5

Multiply $- 5$ by $17$ .

$\frac{2 x + 5 - 85}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.6

Subtract $85$ from $5$ .

$\frac{2 x - 80}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.7

Factor $2$ out of $2 x - 80$ .

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Step 1.7.1

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 80}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.7.2

Factor $2$ out of $- 80$ .

$\frac{2 x + 2 \cdot - 40}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.7.3

Factor $2$ out of $2 x + 2 \cdot - 40$ .

$\frac{2 (x - 40)}{17} + y = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.8

To write $y$ as a fraction with a common denominator, multiply by $\frac{17}{17}$ .

$\frac{2 (x - 40)}{17} + y \cdot \frac{17}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.9

Combine $y$ and $\frac{17}{17}$ .

$\frac{2 (x - 40)}{17} + \frac{y \cdot 17}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.10

Combine the numerators over the common denominator.

$\frac{2 (x - 40) + y \cdot 17}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.11

Simplify the numerator.

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Step 1.11.1

Apply the distributive property.

$\frac{2 x + 2 \cdot - 40 + y \cdot 17}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.11.2

Multiply $2$ by $- 40$ .

$\frac{2 x - 80 + y \cdot 17}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 1.11.3

Move $17$ to the left of $y$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - (1 - x) = 40$

Step 2

Simplify.

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - 1 \cdot 1 - - x = 40$

Step 2.1.2

Multiply $- 1$ by $1$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - 1 - - x = 40$

Step 2.1.3

Multiply $- - x$ .

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Step 2.1.3.1

Multiply $- 1$ by $- 1$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - 1 + 1 x = 40$

Step 2.1.3.2

Multiply $x$ by $1$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - 1 + x = 40$

Step 2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} - 1 \cdot \frac{2}{2} + x = 40$

Step 2.3

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

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62}{2} + \frac{- 1 \cdot 2}{2} + x = 40$

Step 2.4

Combine the numerators over the common denominator.

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62 - 1 \cdot 2}{2} + x = 40$

Step 2.5

Multiply $- 1$ by $2$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 62 - 2}{2} + x = 40$

Step 2.6

Subtract $2$ from $62$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 60}{2} + x = 40$

Step 2.7

To write $x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 60}{2} + x \cdot \frac{2}{2} = 40$

Step 2.8

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

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 60}{2} + \frac{x \cdot 2}{2} = 40$

Step 2.9

Combine the numerators over the common denominator.

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 60 + x \cdot 2}{2} = 40$

Step 2.10

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

$\frac{2 x - 80 + 17 y}{17} = 60, \frac{y + 60 + 2 x}{2} = 40$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{17} & 0 & 60 \\ 0 & \frac{1}{2} & 40 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $17$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $17$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 17 (\frac{1}{17}) & 17 \cdot 0 & 17 \cdot 60 \\ 0 & \frac{1}{2} & 40 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 1020 \\ 0 & \frac{1}{2} & 40 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & 0 & 1020 \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 \cdot 40 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 1020 \\ 0 & 1 & 80 \end{array}]$

Step 5

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

$x = 1020$

$y = 80$

Step 6

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

$(1020, 80)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1020 \\ 80 \end{array}]$