Algebra Examples

$- 2 (x - y) = x - 2 y - 7$ , $6 y = - 30 - 2 x$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Move all terms containing variables to the left side of the equation.

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Step 1.1.1

Subtract $x$ from both sides of the equation.

$- 2 (x - y) - x = - 2 y - 7$

$6 y = - 30 - 2 x$

Step 1.1.2

Add $2 y$ to both sides of the equation.

$- 2 (x - y) - x + 2 y = - 7$

$6 y = - 30 - 2 x$

$- 2 (x - y) - x + 2 y = - 7$

$6 y = - 30 - 2 x$

Step 1.2

Simplify each term.

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Step 1.2.1

Apply the distributive property.

$- 2 x - 2 (- y) - x + 2 y = - 7$

$6 y = - 30 - 2 x$

Step 1.2.2

Multiply $- 1$ by $- 2$ .

$- 2 x + 2 y - x + 2 y = - 7$

$6 y = - 30 - 2 x$

$- 2 x + 2 y - x + 2 y = - 7$

$6 y = - 30 - 2 x$

Step 1.3

Subtract $x$ from $- 2 x$ .

$- 3 x + 2 y + 2 y = - 7$

$6 y = - 30 - 2 x$

Step 1.4

Add $2 y$ and $2 y$ .

$- 3 x + 4 y = - 7$

$6 y = - 30 - 2 x$

Step 1.5

Add $2 x$ to both sides of the equation.

$- 3 x + 4 y = - 7$

$6 y + 2 x = - 30$

Step 1.6

Reorder $6 y$ and $2 x$ .

$- 3 x + 4 y = - 7$

$2 x + 6 y = - 30$

$- 3 x + 4 y = - 7$

$2 x + 6 y = - 30$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & 4 & - 7 \\ 2 & 6 & - 30 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 4 & - \frac{1}{3} \cdot - 7 \\ 2 & 6 & - 30 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} \\ 2 & 6 & - 30 \end{array}]$

Step 3.2

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 3.2.1

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

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} \\ 2 - 2 \cdot 1 & 6 - 2 (- \frac{4}{3}) & - 30 - 2 (\frac{7}{3}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} \\ 0 & \frac{26}{3} & - \frac{104}{3} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{3}{26}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{26}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} \\ \frac{3}{26} \cdot 0 & \frac{3}{26} \cdot \frac{26}{3} & \frac{3}{26} (- \frac{104}{3}) \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{7}{3} \\ 0 & 1 & - 4 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & \frac{7}{3} + \frac{4}{3} \cdot - 4 \\ 0 & 1 & - 4 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x = - 3$

$y = - 4$

Step 5

The solution is the set of ordered pairs that make the system true.

$(- 3, - 4)$