Examples

2 x - 5 y = 3

$2 x - 5 y = 3$ ,

x + 4 > 2 y

$x + 4 > 2 y$

Step 1

Introduce slack variables

u

$u$ and

v

$v$ to replace the inequalities with equations.

x + 4 - Z = 2 y

$x + 4 - Z = 2 y$

2 x - 5 y - 3 = 0

$2 x - 5 y - 3 = 0$

Step 2

Subtract

4

$4$ from both sides of the equation.

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

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

Step 3

Subtract

2 y

$2 y$ from both sides of the equation.

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

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

Step 4

Add

3

$3$ to both sides of the equation.

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

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

Step 5

Write the system of equations in matrix form.

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

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

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}

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

2, 1

$2, 1$ a

0

$0$ .

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Step 6.1.1

Perform the row operation

R_{2} = R_{2} - 2 R_{1}

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

2, 1

$2, 1$ a

0

$0$ .

[\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}]

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

Step 6.1.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 6.2

Multiply each element of

R_{2}

$R_{2}$ by

- 1

$- 1$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 6.2.1

Multiply each element of

R_{2}

$R_{2}$ by

- 1

$- 1$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Step 6.2.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 6.3

Perform the row operation

R_{1} = R_{1} + 2 R_{2}

$R_{1} = R_{1} + 2 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 6.3.1

Perform the row operation

R_{1} = R_{1} + 2 R_{2}

$R_{1} = R_{1} + 2 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

[\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}]

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

Step 6.3.2

Simplify

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Step 7

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

x = 0

$x = 0$

y = 0

$y = 0$

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