Algebra Examples

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

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

$x + 4 - Z = 2 y$

$2 x - 5 y - 3 = 0$

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

$x + 4 - Z - 2 y = 0$

$2 x - 5 y - 3 = 0$

Subtract $4$ from both sides of the equation.

$x - Z - 2 y = - 4$

$2 x - 5 y - 3 = 0$

Add $3$ to both sides of the equation.

$x - Z - 2 y = - 4$

$2 x - 5 y = 3$

Write the system of equations in matrix form.

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

Row reduce.

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Perform the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ .

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

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{2} = - 1 \cdot R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 1 \cdot R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{2} = - 1 \cdot R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 1 \cdot R_{2}$ .

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

$R_{2} = - 1 \cdot R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{1} = 2 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = 2 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{1} = 2 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = 2 \cdot R_{2} + R_{1}$ .

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

$R_{1} = 2 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

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

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

$x = 0$

$y = 0$

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