Linear Algebra Examples

$4 x + 7 y = 10$ , $x + 1.75 y = 2.5$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & 7 & 10 \\ 1 & 1.75 & 2.5 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{7}{4} & \frac{10}{4} \\ 1 & 1.75 & 2.5 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{4} & \frac{5}{2} \\ 1 & 1.75 & 2.5 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{7}{4} & \frac{5}{2} \\ 1 - 1 & 1.75 - \frac{7}{4} & 2.5 - \frac{5}{2} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{4} & \frac{5}{2} \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x + \frac{7 y}{4} = \frac{5}{2}$

Step 4

Solve the equation for $y$ .

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Step 4.1

Subtract $x$ from both sides of the equation.

$\frac{7 y}{4} = \frac{5}{2} - x$

Step 4.2

Multiply both sides of the equation by $\frac{4}{7}$ .

$\frac{4}{7} \cdot \frac{7 y}{4} = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3

Simplify both sides of the equation.

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Step 4.3.1

Simplify the left side.

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Step 4.3.1.1

Simplify $\frac{4}{7} \cdot \frac{7 y}{4}$ .

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Step 4.3.1.1.1

Cancel the common factor of $4$ .

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Step 4.3.1.1.1.1

Cancel the common factor.

$\frac{4}{7} \cdot \frac{7 y}{4} = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3.1.1.1.2

Rewrite the expression.

$\frac{1}{7} (7 y) = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3.1.1.2

Cancel the common factor of $7$ .

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Step 4.3.1.1.2.1

Factor $7$ out of $7 y$ .

$\frac{1}{7} (7 (y)) = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{1}{7} (7 y) = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3.1.1.2.3

Rewrite the expression.

$y = \frac{4}{7} (\frac{5}{2} - x)$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Simplify $\frac{4}{7} (\frac{5}{2} - x)$ .

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Step 4.3.2.1.1

Apply the distributive property.

$y = \frac{4}{7} \cdot \frac{5}{2} + \frac{4}{7} (- x)$

Step 4.3.2.1.2

Cancel the common factor of $2$ .

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Step 4.3.2.1.2.1

Factor $2$ out of $4$ .

$y = \frac{2 (2)}{7} \cdot \frac{5}{2} + \frac{4}{7} (- x)$

Step 4.3.2.1.2.2

Cancel the common factor.

$y = \frac{2 \cdot 2}{7} \cdot \frac{5}{2} + \frac{4}{7} (- x)$

Step 4.3.2.1.2.3

Rewrite the expression.

$y = \frac{2}{7} \cdot 5 + \frac{4}{7} (- x)$

Step 4.3.2.1.3

Combine $\frac{2}{7}$ and $5$ .

$y = \frac{2 \cdot 5}{7} + \frac{4}{7} (- x)$

Step 4.3.2.1.4

Multiply $2$ by $5$ .

$y = \frac{10}{7} + \frac{4}{7} (- x)$

Step 4.3.2.1.5

Combine $\frac{4}{7}$ and $x$ .

$y = \frac{10}{7} - \frac{4 x}{7}$

Step 4.4

Reorder $\frac{10}{7}$ and $- \frac{4 x}{7}$ .

$y = - \frac{4 x}{7} + \frac{10}{7}$

Step 5

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

$(x, - \frac{4 x}{7} + \frac{10}{7})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} x \\ - \frac{4 x}{7} + \frac{10}{7} \end{array}]$