Linear Algebra Examples

$y = 4 x + 3 x - 2$ , $y = 6$

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 $4 x$ from both sides of the equation.

$y - 4 x = 3 x - 2$

$y = 6$

Step 1.1.2

Subtract $3 x$ from both sides of the equation.

$y - 4 x - 3 x = - 2$

$y = 6$

$y - 4 x - 3 x = - 2$

$y = 6$

Step 1.2

Subtract $3 x$ from $- 4 x$ .

$y - 7 x = - 2$

$y = 6$

Step 1.3

Reorder $y$ and $- 7 x$ .

$- 7 x + y = - 2$

$y = 6$

$- 7 x + y = - 2$

$y = 6$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 7 & 1 & - 2 \\ 0 & 1 & 6 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 1 & - \frac{1}{7} \cdot - 2 \\ 0 & 1 & 6 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & \frac{2}{7} \\ 0 & 1 & 6 \end{array}]$

Step 3.2

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{7} \cdot 0 & - \frac{1}{7} + \frac{1}{7} \cdot 1 & \frac{2}{7} + \frac{1}{7} \cdot 6 \\ 0 & 1 & 6 \end{array}]$

Step 3.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{7} \\ 0 & 1 & 6 \end{array}]$

Step 4

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

$x = \frac{8}{7}$

$y = 6$

Step 5

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

$(\frac{8}{7}, 6)$