Algebra Examples

$2 x + 2 y = 6$ , $x + 2 y > 9$

Step 1

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

$x + 2 y - Z = 9$

$2 x + 2 y - 6 = 0$

Step 2

Add $6$ to both sides of the equation.

$2 x + 2 y = 6$

$x + 2 y - Z = 9$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ 2 & 2 & 0 & 6 \end{array}]$

Step 4

Find the reduced row echelon form of the matrix.

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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 $0$ .

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (2) + 2 & (- 2) \cdot (0) + 0 & (- 2) \cdot (9) + 6 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ 0 & - 2 & 0 & - 12 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{2} R_{2}$ 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 $R_{2} = - \frac{1}{2} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ (- \frac{1}{2}) \cdot (0) & (- \frac{1}{2}) \cdot (- 2) & (- \frac{1}{2}) \cdot (0) & (- \frac{1}{2}) \cdot (- 12) \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 0 & 9 \\ 0 & 1 & 0 & 6 \end{array}]$

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 $0$ .

$[\begin{array}{c} - 2 \cdot R_{2} + R_{1} & - 2 \cdot R_{2} + R_{1} & - 2 \cdot R_{2} + R_{1} & - 2 \cdot R_{2} + R_{1} \\ 0 & 1 & 0 & 6 \end{array}]$

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

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

$[\begin{array}{c} (- 2) \cdot (0) + 1 & (- 2) \cdot (1) + 2 & (- 2) \cdot (0) + 0 & (- 2) \cdot (6) + 9 \\ 0 & 1 & 0 & 6 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & - 3 \\ 0 & 1 & 0 & 6 \end{array}]$

Step 5

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

$x = 0$

$y = 6$

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