Precalculus Examples

$x + 2 = 5$ , $- 2 x + y = 5$

Step 1

Move all terms not containing a variable to the right side of the equation.

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Step 1.1

Subtract $2$ from both sides of the equation.

$x = 5 - 2$

$- 2 x + y = 5$

Step 1.2

Subtract $2$ from $5$ .

$x = 3$

$- 2 x + y = 5$

$x = 3$

$- 2 x + y = 5$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 0 & 3 \\ - 2 & 1 & 5 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 0 & 3 \\ - 2 + 2 \cdot 1 & 1 + 2 \cdot 0 & 5 + 2 \cdot 3 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 3 \\ 0 & 1 & 11 \end{array}]$

Step 4

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

$x = 3$

$y = 11$

Step 5

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

$(3, 11)$