Examples

$2 x - 5 y = 3$ , $x + 4 > 2 y$

Step 1

Introduce slack variables $u$ and $v$ to replace the inequalities with equations.

$x + 4 - Z = 2 y$

$2 x - 5 y - 3 = 0$

Step 2

Subtract $4$ from both sides of the equation.

$x - Z = 2 y - 4, 2 x - 5 y - 3 = 0$

Step 3

Subtract $2 y$ from both sides of the equation.

$x - Z - 2 y = - 4, 2 x - 5 y - 3 = 0$

Step 4

Add $3$ to both sides of the equation.

$x - Z - 2 y = - 4, 2 x - 5 y = 3$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 2 & 0 & - 4 \\ 2 & - 5 & 0 & 3 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

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 6.1.1

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

$[\begin{array}{c} 1 & - 2 & 0 & - 4 \\ 2 - 2 \cdot 1 & - 5 - 2 \cdot - 2 & 0 - 2 \cdot 0 & 3 - 2 \cdot - 4 \end{array}]$

Step 6.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - 4 \\ 0 & - 1 & 0 & 11 \end{array}]$

Step 6.2

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

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Step 6.2.1

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

$[\begin{array}{c} 1 & - 2 & 0 & - 4 \\ - 0 & - - 1 & - 0 & - 1 \cdot 11 \end{array}]$

Step 6.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - 4 \\ 0 & 1 & 0 & - 11 \end{array}]$

Step 6.3

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

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Step 6.3.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 0 + 2 \cdot 0 & - 4 + 2 \cdot - 11 \\ 0 & 1 & 0 & - 11 \end{array}]$

Step 6.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 26 \\ 0 & 1 & 0 & - 11 \end{array}]$

Step 7

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

$x = 0$

$y = 0$

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