Algebra Examples

$x - 2 y = 2$ $y = - \frac{1}{2} x + 5$

Step 1

Add $2 y$ to both sides of the equation.

$x = 2 + 2 y$

$y = - \frac{1}{2} x + 5$

Step 2

Replace all occurrences of $x$ with $2 + 2 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $y = - \frac{1}{2} x + 5$ with $2 + 2 y$ .

$y = - \frac{1}{2} \cdot (2 + 2 y) + 5$

$x = 2 + 2 y$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $- \frac{1}{2} \cdot (2 + 2 y) + 5$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$y = - \frac{1}{2} \cdot 2 - \frac{1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y = \frac{- 1}{2} \cdot 2 - \frac{1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.2.2

Cancel the common factor.

$y = \frac{- 1}{2} \cdot 2 - \frac{1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.2.3

Rewrite the expression.

$y = - 1 - \frac{1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

$y = - 1 - \frac{1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.3

Cancel the common factor of $2$ .

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Step 2.2.1.1.3.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y = - 1 + \frac{- 1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.3.2

Factor $2$ out of $2 y$ .

$y = - 1 + \frac{- 1}{2} \cdot (2 (y)) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.3.3

Cancel the common factor.

$y = - 1 + \frac{- 1}{2} \cdot (2 y) + 5$

$x = 2 + 2 y$

Step 2.2.1.1.3.4

Rewrite the expression.

$y = - 1 - 1 y + 5$

$x = 2 + 2 y$

$y = - 1 - 1 y + 5$

$x = 2 + 2 y$

Step 2.2.1.1.4

Rewrite $- 1 y$ as $- y$ .

$y = - 1 - y + 5$

$x = 2 + 2 y$

$y = - 1 - y + 5$

$x = 2 + 2 y$

Step 2.2.1.2

Add $- 1$ and $5$ .

$y = - y + 4$

$x = 2 + 2 y$

$y = - y + 4$

$x = 2 + 2 y$

$y = - y + 4$

$x = 2 + 2 y$

$y = - y + 4$

$x = 2 + 2 y$

Step 3

Solve for $y$ in $y = - y + 4$ .

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Step 3.1

Move all terms containing $y$ to the left side of the equation.

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Step 3.1.1

Add $y$ to both sides of the equation.

$y + y = 4$

$x = 2 + 2 y$

Step 3.1.2

Add $y$ and $y$ .

$2 y = 4$

$x = 2 + 2 y$

$2 y = 4$

$x = 2 + 2 y$

Step 3.2

Divide each term in $2 y = 4$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 y = 4$ by $2$ .

$\frac{2 y}{2} = \frac{4}{2}$

$x = 2 + 2 y$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{4}{2}$

$x = 2 + 2 y$

Step 3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{4}{2}$

$x = 2 + 2 y$

$y = \frac{4}{2}$

$x = 2 + 2 y$

$y = \frac{4}{2}$

$x = 2 + 2 y$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $4$ by $2$ .

$y = 2$

$x = 2 + 2 y$

$y = 2$

$x = 2 + 2 y$

$y = 2$

$x = 2 + 2 y$

$y = 2$

$x = 2 + 2 y$

Step 4

Replace all occurrences of $y$ with $2$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 2 + 2 y$ with $2$ .

$x = 2 + 2 (2)$

$y = 2$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $2 + 2 (2)$ .

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Step 4.2.1.1

Multiply $2$ by $2$ .

$x = 2 + 4$

$y = 2$

Step 4.2.1.2

Add $2$ and $4$ .

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(6, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, 2)$

Equation Form:

$x = 6, y = 2$

Step 7