Linear Algebra Examples

$\frac{x + 5}{2} - \frac{y + 5}{3} = 4$ , $\frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1

Simplify.

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Step 1.1

To write $\frac{x + 5}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x + 5}{2} \cdot \frac{3}{3} - \frac{y + 5}{3} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.2

To write $- \frac{y + 5}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{x + 5}{2} \cdot \frac{3}{3} - \frac{y + 5}{3} \cdot \frac{2}{2} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.3

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{x + 5}{2}$ by $\frac{3}{3}$ .

$\frac{(x + 5) \cdot 3}{2 \cdot 3} - \frac{y + 5}{3} \cdot \frac{2}{2} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.3.2

Multiply $2$ by $3$ .

$\frac{(x + 5) \cdot 3}{6} - \frac{y + 5}{3} \cdot \frac{2}{2} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.3.3

Multiply $\frac{y + 5}{3}$ by $\frac{2}{2}$ .

$\frac{(x + 5) \cdot 3}{6} - \frac{(y + 5) \cdot 2}{3 \cdot 2} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.3.4

Multiply $3$ by $2$ .

$\frac{(x + 5) \cdot 3}{6} - \frac{(y + 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.4

Combine the numerators over the common denominator.

$\frac{(x + 5) \cdot 3 - (y + 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5

Simplify the numerator.

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Step 1.5.1

Apply the distributive property.

$\frac{x \cdot 3 + 5 \cdot 3 - (y + 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.2

Move $3$ to the left of $x$ .

$\frac{3 \cdot x + 5 \cdot 3 - (y + 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.3

Multiply $5$ by $3$ .

$\frac{3 \cdot x + 15 - (y + 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.4

Apply the distributive property.

$\frac{3 x + 15 + (- y - 1 \cdot 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.5

Multiply $- 1$ by $5$ .

$\frac{3 x + 15 + (- y - 5) \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.6

Apply the distributive property.

$\frac{3 x + 15 - y \cdot 2 - 5 \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.7

Multiply $2$ by $- 1$ .

$\frac{3 x + 15 - 2 y - 5 \cdot 2}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.8

Multiply $- 5$ by $2$ .

$\frac{3 x + 15 - 2 y - 10}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 1.5.9

Subtract $10$ from $15$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + y}{2} = \frac{1}{3} + \frac{x - y}{3}$

Step 2

Combine the numerators over the common denominator.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + y}{2} = \frac{1 + x - y}{3}$

Step 3

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

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Step 3.1

Subtract $\frac{1 + x - y}{3}$ from both sides of the equation.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + y}{2} - \frac{1 + x - y}{3} = 0$

Step 3.2

To write $\frac{x + y}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + y}{2} \cdot \frac{3}{3} - \frac{1 + x - y}{3} = 0$

Step 3.3

To write $- \frac{1 + x - y}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + y}{2} \cdot \frac{3}{3} - \frac{1 + x - y}{3} \cdot \frac{2}{2} = 0$

Step 3.4

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.4.1

Multiply $\frac{x + y}{2}$ by $\frac{3}{3}$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{(x + y) \cdot 3}{2 \cdot 3} - \frac{1 + x - y}{3} \cdot \frac{2}{2} = 0$

Step 3.4.2

Multiply $2$ by $3$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{(x + y) \cdot 3}{6} - \frac{1 + x - y}{3} \cdot \frac{2}{2} = 0$

Step 3.4.3

Multiply $\frac{1 + x - y}{3}$ by $\frac{2}{2}$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{(x + y) \cdot 3}{6} - \frac{(1 + x - y) \cdot 2}{3 \cdot 2} = 0$

Step 3.4.4

Multiply $3$ by $2$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{(x + y) \cdot 3}{6} - \frac{(1 + x - y) \cdot 2}{6} = 0$

Step 3.5

Combine the numerators over the common denominator.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{(x + y) \cdot 3 - (1 + x - y) \cdot 2}{6} = 0$

Step 3.6

Simplify the numerator.

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Step 3.6.1

Apply the distributive property.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x \cdot 3 + y \cdot 3 - (1 + x - y) \cdot 2}{6} = 0$

Step 3.6.2

Move $3$ to the left of $x$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 \cdot x + y \cdot 3 - (1 + x - y) \cdot 2}{6} = 0$

Step 3.6.3

Move $3$ to the left of $y$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 \cdot x + 3 \cdot y - (1 + x - y) \cdot 2}{6} = 0$

Step 3.6.4

Multiply $3$ by $y$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y - (1 + x - y) \cdot 2}{6} = 0$

Step 3.6.5

Apply the distributive property.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y + (- 1 \cdot 1 - x - - y) \cdot 2}{6} = 0$

Step 3.6.6

Simplify.

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Step 3.6.6.1

Multiply $- 1$ by $1$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y + (- 1 - x - - y) \cdot 2}{6} = 0$

Step 3.6.6.2

Multiply $- - y$ .

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Step 3.6.6.2.1

Multiply $- 1$ by $- 1$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y + (- 1 - x + 1 y) \cdot 2}{6} = 0$

Step 3.6.6.2.2

Multiply $y$ by $1$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y + (- 1 - x + y) \cdot 2}{6} = 0$

Step 3.6.7

Apply the distributive property.

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y - 1 \cdot 2 - x \cdot 2 + y \cdot 2}{6} = 0$

Step 3.6.8

Simplify.

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Step 3.6.8.1

Multiply $- 1$ by $2$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y - 2 - x \cdot 2 + y \cdot 2}{6} = 0$

Step 3.6.8.2

Multiply $2$ by $- 1$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y - 2 - 2 x + y \cdot 2}{6} = 0$

Step 3.6.8.3

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

$\frac{3 x - 2 y + 5}{6} = 4, \frac{3 x + 3 y - 2 - 2 x + 2 \cdot y}{6} = 0$

Step 3.6.9

Subtract $2 x$ from $3 x$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + 3 y - 2 + 2 y}{6} = 0$

Step 3.6.10

Add $3 y$ and $2 y$ .

$\frac{3 x - 2 y + 5}{6} = 4, \frac{x + 5 y - 2}{6} = 0$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{6} & 0 & 4 \\ \frac{1}{6} & 0 & 0 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Multiply each element of $R_{1}$ by $6$ to make the entry at $1, 1$ a $1$ .

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Step 5.1.1

Multiply each element of $R_{1}$ by $6$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 6 (\frac{1}{6}) & 6 \cdot 0 & 6 \cdot 4 \\ \frac{1}{6} & 0 & 0 \end{array}]$

Step 5.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 24 \\ \frac{1}{6} & 0 & 0 \end{array}]$

Step 5.2

Perform the row operation $R_{2} = R_{2} - \frac{1}{6} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 5.2.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{6} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 24 \\ \frac{1}{6} - \frac{1}{6} \cdot 1 & 0 - \frac{1}{6} \cdot 0 & 0 - \frac{1}{6} \cdot 24 \end{array}]$

Step 5.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 24 \\ 0 & 0 & - 4 \end{array}]$

Step 5.3

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 3$ a $1$ .

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Step 5.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & 24 \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 \end{array}]$

Step 5.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 24 \\ 0 & 0 & 1 \end{array}]$

Step 5.4

Perform the row operation $R_{1} = R_{1} - 24 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 5.4.1

Perform the row operation $R_{1} = R_{1} - 24 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 24 \cdot 0 & 0 - 24 \cdot 0 & 24 - 24 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 5.4.2

Simplify $R_{1}$ .

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

Step 6

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

$x = 0$

$0 = 1$

Step 7

Since $0 \neq 1$ , there are no solutions.

No solution