Algebra Examples

$x - y + z = - 1$

Step 1

Solve the equation for $y$ .

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Step 1.1

Move all terms not containing $y$ to the right side of the equation.

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Step 1.1.1

Subtract $x$ from both sides of the equation.

$- y + z = - 1 - x$

Step 1.1.2

Subtract $z$ from both sides of the equation.

$- y = - 1 - x - z$

Step 1.2

Divide each term in $- y = - 1 - x - z$ by $- 1$ and simplify.

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Step 1.2.1

Divide each term in $- y = - 1 - x - z$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 1}{- 1} + \frac{- x}{- 1} + \frac{- z}{- 1}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 1}{- 1} + \frac{- x}{- 1} + \frac{- z}{- 1}$

Step 1.2.2.2

Divide $y$ by $1$ .

$y = \frac{- 1}{- 1} + \frac{- x}{- 1} + \frac{- z}{- 1}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $- 1$ by $- 1$ .

$y = 1 + \frac{- x}{- 1} + \frac{- z}{- 1}$

Step 1.2.3.1.2

Dividing two negative values results in a positive value.

$y = 1 + \frac{x}{1} + \frac{- z}{- 1}$

Step 1.2.3.1.3

Divide $x$ by $1$ .

$y = 1 + x + \frac{- z}{- 1}$

Step 1.2.3.1.4

Dividing two negative values results in a positive value.

$y = 1 + x + \frac{z}{1}$

Step 1.2.3.1.5

Divide $z$ by $1$ .

$y = 1 + x + z$

Step 2

Choose any values for $x$ and $y$ that are in the domain to plug into the equation.

Step 3

Choose $0$ to substitute in for $x$ and $1$ to substitute for $z$ .

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Step 3.1

Remove parentheses.

$y = 1 + 0 + 1$

Step 3.2

Simplify $1 + 0 + 1$ .

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Step 3.2.1

Add $1$ and $0$ .

$y = 1 + 1$

Step 3.2.2

Add $1$ and $1$ .

$y = 2$

Step 3.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(0, 2, 1)$

Step 4

Choose $1$ to substitute in for $x$ and $2$ to substitute for $z$ .

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Step 4.1

Remove parentheses.

$y = 1 + 1 + 2$

Step 4.2

Simplify $1 + 1 + 2$ .

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Step 4.2.1

Add $1$ and $1$ .

$y = 2 + 2$

Step 4.2.2

Add $2$ and $2$ .

$y = 4$

Step 4.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(1, 4, 2)$

Step 5

Choose $2$ to substitute in for $x$ and $3$ to substitute for $z$ .

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Step 5.1

Remove parentheses.

$y = 1 + 2 + 3$

Step 5.2

Simplify $1 + 2 + 3$ .

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Step 5.2.1

Add $1$ and $2$ .

$y = 3 + 3$

Step 5.2.2

Add $3$ and $3$ .

$y = 6$

Step 5.3

Use the $x$ , $y$ , and $z$ values to form the ordered pair.

$(2, 6, 3)$

Step 6

These are three possible solutions to the equation.

$(0, 2, 1), (1, 4, 2), (2, 6, 3)$