Algebra Examples

$3 (x + 2) = 7 y$ , $x + y > - 1$

Step 1

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

$x + y - Z = - 1$

$3 x + 6 - 7 y = 0$

Step 2

Subtract $6$ from both sides of the equation.

$3 x - 7 y = - 6$

$x + y - Z = - 1$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ 3 & - 7 & 0 & - 6 \end{array}]$

Step 4

Find the reduced row echelon form of the matrix.

Tap for more steps...

Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Tap for more steps...

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (1) - 7 & (- 3) \cdot (0) + 0 & (- 3) \cdot (- 1) - 6 \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ 0 & - 10 & 0 & - 3 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{10} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Tap for more steps...

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - \frac{1}{10} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} & - \frac{1}{10} R_{2} \end{array}]$

$R_{2} = - \frac{1}{10} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{10} R_{2}$ .

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ (- \frac{1}{10}) \cdot (0) & (- \frac{1}{10}) \cdot (- 10) & (- \frac{1}{10}) \cdot (0) & (- \frac{1}{10}) \cdot (- 3) \end{array}]$

$R_{2} = - \frac{1}{10} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 0 & - 1 \\ 0 & 1 & 0 & \frac{3}{10} \end{array}]$

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

Tap for more steps...

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

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & 0 & \frac{3}{10} \end{array}]$

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (0) + 0 & (- 1) \cdot (\frac{3}{10}) - 1 \\ 0 & 1 & 0 & \frac{3}{10} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & - \frac{13}{10} \\ 0 & 1 & 0 & \frac{3}{10} \end{array}]$

Step 5

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

$x = 0$

$y = \frac{3}{10}$

Enter YOUR Problem