Linear Algebra Examples

$1575 = (x + y) \cdot 4.5$ , $1575 = (x + y) \cdot 3.5$

Step 1

Simplify.

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Step 1.1

Apply the distributive property.

$1575 = x \cdot 4.5 + y \cdot 4.5, 1575 = (x + y) \cdot 3.5$

Step 1.2

Move $4.5$ to the left of $x$ .

$1575 = 4.5 \cdot x + y \cdot 4.5, 1575 = (x + y) \cdot 3.5$

Step 1.3

Move $4.5$ to the left of $y$ .

$1575 = 4.5 \cdot x + 4.5 \cdot y, 1575 = (x + y) \cdot 3.5$

Step 1.4

Multiply $4.5$ by $y$ .

$1575 = 4.5 x + 4.5 y, 1575 = (x + y) \cdot 3.5$

Step 2

Subtract $1575$ from both sides of the equation.

$0 = 4.5 x + 4.5 y - 1575, 1575 = (x + y) \cdot 3.5$

Step 3

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

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Step 3.1

Subtract $4.5 x$ from both sides of the equation.

$- 4.5 x = 4.5 y - 1575, 1575 = (x + y) \cdot 3.5$

Step 3.2

Subtract $4.5 y$ from both sides of the equation.

$- 4.5 x - 4.5 y = - 1575, 1575 = (x + y) \cdot 3.5$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$- 4.5 x - 4.5 y = - 1575, 1575 = x \cdot 3.5 + y \cdot 3.5$

Step 4.2

Move $3.5$ to the left of $x$ .

$- 4.5 x - 4.5 y = - 1575, 1575 = 3.5 \cdot x + y \cdot 3.5$

Step 4.3

Move $3.5$ to the left of $y$ .

$- 4.5 x - 4.5 y = - 1575, 1575 = 3.5 \cdot x + 3.5 \cdot y$

Step 4.4

Multiply $3.5$ by $y$ .

$- 4.5 x - 4.5 y = - 1575, 1575 = 3.5 x + 3.5 y$

Step 5

Subtract $1575$ from both sides of the equation.

$- 4.5 x - 4.5 y = - 1575, 0 = 3.5 x + 3.5 y - 1575$

Step 6

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

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Step 6.1

Subtract $3.5 x$ from both sides of the equation.

$- 4.5 x - 4.5 y = - 1575, - 3.5 x = 3.5 y - 1575$

Step 6.2

Subtract $3.5 y$ from both sides of the equation.

$- 4.5 x - 4.5 y = - 1575, - 3.5 x - 3.5 y = - 1575$

Step 7

Write the system of equations in matrix form.

$[\begin{array}{c} - 4.5 & - 4.5 & - 1575 \\ - 3.5 & - 3.5 & - 1575 \end{array}]$

Step 8

Find the reduced row echelon form.

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Step 8.1

Multiply each element of $R_{1}$ by $\frac{1}{- 4.5}$ to make the entry at $1, 1$ a $1$ .

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Step 8.1.1

Multiply each element of $R_{1}$ by $\frac{1}{- 4.5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{- 4.5}{- 4.5} & \frac{- 4.5}{- 4.5} & \frac{- 1575}{- 4.5} \\ - 3.5 & - 3.5 & - 1575 \end{array}]$

Step 8.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 350 \\ - 3.5 & - 3.5 & - 1575 \end{array}]$

Step 8.2

Perform the row operation $R_{2} = R_{2} + 3.5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 8.2.1

Perform the row operation $R_{2} = R_{2} + 3.5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 350 \\ - 3.5 + 3.5 \cdot 1 & - 3.5 + 3.5 \cdot 1 & - 1575 + 3.5 \cdot 350 \end{array}]$

Step 8.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 350 \\ 0 & 0 & - 350 \end{array}]$

Step 8.3

Multiply each element of $R_{2}$ by $- \frac{1}{350}$ to make the entry at $2, 3$ a $1$ .

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Step 8.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{350}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 350 \\ - \frac{1}{350} \cdot 0 & - \frac{1}{350} \cdot 0 & - \frac{1}{350} \cdot - 350 \end{array}]$

Step 8.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 350 \\ 0 & 0 & 1 \end{array}]$

Step 8.4

Perform the row operation $R_{1} = R_{1} - 350 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 8.4.1

Perform the row operation $R_{1} = R_{1} - 350 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 350 \cdot 0 & 1 - 350 \cdot 0 & 350 - 350 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 8.4.2

Simplify $R_{1}$ .

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

Step 9

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

$x + y = 0$

$0 = 1$

Step 10

Subtract $y$ from both sides of the equation.

$x = - y$

$0 = 1$

Step 11

Since $0 \neq 1$ , there are no solutions.

No solution