Linear Algebra Examples

$4 x + 5 y = 3$ , $8 x + 10 y = 6$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & 5 & 3 \\ 8 & 10 & 6 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

Tap for more steps...

Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{5}{4} & \frac{3}{4} \\ 8 & 10 & 6 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{3}{4} \\ 8 & 10 & 6 \end{array}]$

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{5}{4} & \frac{3}{4} \\ 8 - 8 \cdot 1 & 10 - 8 (\frac{5}{4}) & 6 - 8 (\frac{3}{4}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

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

Step 3

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

$x + \frac{5 y}{4} = \frac{3}{4}$

Step 4

Solve the equation for $y$ .

Tap for more steps...

Step 4.1

Subtract $x$ from both sides of the equation.

$\frac{5 y}{4} = \frac{3}{4} - x$

Step 4.2

Multiply both sides of the equation by $\frac{4}{5}$ .

$\frac{4}{5} \cdot \frac{5 y}{4} = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3

Simplify both sides of the equation.

Tap for more steps...

Step 4.3.1

Simplify the left side.

Tap for more steps...

Step 4.3.1.1

Simplify $\frac{4}{5} \cdot \frac{5 y}{4}$ .

Tap for more steps...

Step 4.3.1.1.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 4.3.1.1.1.1

Cancel the common factor.

$\frac{4}{5} \cdot \frac{5 y}{4} = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3.1.1.1.2

Rewrite the expression.

$\frac{1}{5} (5 y) = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3.1.1.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 4.3.1.1.2.1

Factor $5$ out of $5 y$ .

$\frac{1}{5} (5 (y)) = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{1}{5} (5 y) = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3.1.1.2.3

Rewrite the expression.

$y = \frac{4}{5} (\frac{3}{4} - x)$

Step 4.3.2

Simplify the right side.

Tap for more steps...

Step 4.3.2.1

Simplify $\frac{4}{5} (\frac{3}{4} - x)$ .

Tap for more steps...

Step 4.3.2.1.1

Apply the distributive property.

$y = \frac{4}{5} \cdot \frac{3}{4} + \frac{4}{5} (- x)$

Step 4.3.2.1.2

Cancel the common factor of $4$ .

Tap for more steps...

Step 4.3.2.1.2.1

Cancel the common factor.

$y = \frac{4}{5} \cdot \frac{3}{4} + \frac{4}{5} (- x)$

Step 4.3.2.1.2.2

Rewrite the expression.

$y = \frac{1}{5} \cdot 3 + \frac{4}{5} (- x)$

Step 4.3.2.1.3

Combine $\frac{1}{5}$ and $3$ .

$y = \frac{3}{5} + \frac{4}{5} (- x)$

Step 4.3.2.1.4

Combine $\frac{4}{5}$ and $x$ .

$y = \frac{3}{5} - \frac{4 x}{5}$

Step 4.4

Reorder $\frac{3}{5}$ and $- \frac{4 x}{5}$ .

$y = - \frac{4 x}{5} + \frac{3}{5}$

Step 5

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

$(x, - \frac{4 x}{5} + \frac{3}{5})$

Step 6

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ - \frac{4 x}{5} + \frac{3}{5} \end{array}]$