Linear Algebra Examples

$5 x + 3 = 4 y$ , $y = 8 x - 2$

Step 1

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

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Step 1.1

Subtract $4 y$ from both sides of the equation.

$5 x + 3 - 4 y = 0$

$y = 8 x - 2$

Step 1.2

Subtract $3$ from both sides of the equation.

$5 x - 4 y = - 3$

$y = 8 x - 2$

Step 1.3

Subtract $8 x$ from both sides of the equation.

$5 x - 4 y = - 3$

$y - 8 x = - 2$

Step 1.4

Reorder $y$ and $- 8 x$ .

$5 x - 4 y = - 3$

$- 8 x + y = - 2$

$5 x - 4 y = - 3$

$- 8 x + y = - 2$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 5 & - 4 & - 3 \\ - 8 & 1 & - 2 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 4}{5} & \frac{- 3}{5} \\ - 8 & 1 & - 2 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ - 8 & 1 & - 2 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ - 8 + 8 \cdot 1 & 1 + 8 (- \frac{4}{5}) & - 2 + 8 (- \frac{3}{5}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & - \frac{27}{5} & - \frac{34}{5} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{5}{27}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{27}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ - \frac{5}{27} \cdot 0 & - \frac{5}{27} (- \frac{27}{5}) & - \frac{5}{27} (- \frac{34}{5}) \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{5} \cdot 0 & - \frac{4}{5} + \frac{4}{5} \cdot 1 & - \frac{3}{5} + \frac{4}{5} \cdot \frac{34}{27} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{11}{27} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 4

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

$x = \frac{11}{27}$

$y = \frac{34}{27}$

Step 5

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

$(\frac{11}{27}, \frac{34}{27})$