Algebra Examples

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

Step 1

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

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Step 1.1

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

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

$y = 2 x + 2$

Step 1.2

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

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Step 1.2.1

Simplify the left side.

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Step 1.2.1.1

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

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Step 1.2.1.1.1

Apply the distributive property.

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

$y = 2 x + 2$

Step 1.2.1.1.2

Cancel the common factor of $2$ .

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Step 1.2.1.1.2.1

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

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

$y = 2 x + 2$

Step 1.2.1.1.2.2

Factor $2$ out of $2 x$ .

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

$y = 2 x + 2$

Step 1.2.1.1.2.3

Cancel the common factor.

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

$y = 2 x + 2$

Step 1.2.1.1.2.4

Rewrite the expression.

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

$y = 2 x + 2$

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

$y = 2 x + 2$

Step 1.2.1.1.3

Cancel the common factor of $2$ .

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Step 1.2.1.1.3.1

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

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

$y = 2 x + 2$

Step 1.2.1.1.3.2

Cancel the common factor.

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

$y = 2 x + 2$

Step 1.2.1.1.3.3

Rewrite the expression.

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

$y = 2 x + 2$

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

$y = 2 x + 2$

Step 1.2.1.1.4

Rewrite $- 1 x$ as $- x$ .

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

$y = 2 x + 2$

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

$y = 2 x + 2$

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

$y = 2 x + 2$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Combine $\frac{1}{2}$ and $x$ .

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

$y = 2 x + 2$

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

$y = 2 x + 2$

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

$y = 2 x + 2$

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

$y = 2 x + 2$

Step 2

Solve for $x$ in $- x - 1 = \frac{x}{2} + 5$ .

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Step 2.1

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

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Step 2.1.1

Subtract $\frac{x}{2}$ from both sides of the equation.

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

$y = 2 x + 2$

Step 2.1.2

To write $- x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- x \cdot \frac{2}{2} - \frac{x}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.1.3

Combine $- x$ and $\frac{2}{2}$ .

$\frac{- x \cdot 2}{2} - \frac{x}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.1.4

Combine the numerators over the common denominator.

$\frac{- x \cdot 2 - x}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.1.5

Simplify each term.

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Step 2.1.5.1

Simplify the numerator.

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Step 2.1.5.1.1

Factor $x$ out of $- x \cdot 2 - x$ .

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Step 2.1.5.1.1.1

Factor $x$ out of $- x \cdot 2$ .

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

$y = 2 x + 2$

Step 2.1.5.1.1.2

Factor $x$ out of $- x$ .

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

$y = 2 x + 2$

Step 2.1.5.1.1.3

Factor $x$ out of $x (- 1 \cdot 2) + x \cdot - 1$ .

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

$y = 2 x + 2$

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

$y = 2 x + 2$

Step 2.1.5.1.2

Multiply $- 1$ by $2$ .

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

$y = 2 x + 2$

Step 2.1.5.1.3

Subtract $1$ from $- 2$ .

$\frac{x \cdot - 3}{2} - 1 = 5$

$y = 2 x + 2$

$\frac{x \cdot - 3}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.1.5.2

Move $- 3$ to the left of $x$ .

$\frac{- 3 \cdot x}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.1.5.3

Move the negative in front of the fraction.

$- \frac{3 x}{2} - 1 = 5$

$y = 2 x + 2$

$- \frac{3 x}{2} - 1 = 5$

$y = 2 x + 2$

$- \frac{3 x}{2} - 1 = 5$

$y = 2 x + 2$

Step 2.2

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

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Step 2.2.1

Add $1$ to both sides of the equation.

$- \frac{3 x}{2} = 5 + 1$

$y = 2 x + 2$

Step 2.2.2

Add $5$ and $1$ .

$- \frac{3 x}{2} = 6$

$y = 2 x + 2$

$- \frac{3 x}{2} = 6$

$y = 2 x + 2$

Step 2.3

Multiply both sides of the equation by $- \frac{2}{3}$ .

$- \frac{2}{3} \cdot (- \frac{3 x}{2}) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4

Simplify both sides of the equation.

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Step 2.4.1

Simplify the left side.

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Step 2.4.1.1

Simplify $- \frac{2}{3} (- \frac{3 x}{2})$ .

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Step 2.4.1.1.1

Cancel the common factor of $2$ .

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Step 2.4.1.1.1.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$\frac{- 2}{3} \cdot (- \frac{3 x}{2}) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.1.2

Move the leading negative in $- \frac{3 x}{2}$ into the numerator.

$\frac{- 2}{3} \cdot \frac{- 3 x}{2} = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.1.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{3} \cdot \frac{- 3 x}{2} = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.1.4

Cancel the common factor.

$\frac{2 \cdot - 1}{3} \cdot \frac{- 3 x}{2} = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.1.5

Rewrite the expression.

$\frac{- 1}{3} \cdot (- 3 x) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

$\frac{- 1}{3} \cdot (- 3 x) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.2

Cancel the common factor of $3$ .

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Step 2.4.1.1.2.1

Factor $3$ out of $- 3 x$ .

$\frac{- 1}{3} \cdot (3 (- x)) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.2.2

Cancel the common factor.

$\frac{- 1}{3} \cdot (3 (- x)) = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.2.3

Rewrite the expression.

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.3

Multiply.

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Step 2.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.1.1.3.2

Multiply $x$ by $1$ .

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

$x = - \frac{2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify $- \frac{2}{3} \cdot 6$ .

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Step 2.4.2.1.1

Cancel the common factor of $3$ .

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Step 2.4.2.1.1.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$x = \frac{- 2}{3} \cdot 6$

$y = 2 x + 2$

Step 2.4.2.1.1.2

Factor $3$ out of $6$ .

$x = \frac{- 2}{3} \cdot (3 (2))$

$y = 2 x + 2$

Step 2.4.2.1.1.3

Cancel the common factor.

$x = \frac{- 2}{3} \cdot (3 \cdot 2)$

$y = 2 x + 2$

Step 2.4.2.1.1.4

Rewrite the expression.

$x = - 2 \cdot 2$

$y = 2 x + 2$

$x = - 2 \cdot 2$

$y = 2 x + 2$

Step 2.4.2.1.2

Multiply $- 2$ by $2$ .

$x = - 4$

$y = 2 x + 2$

$x = - 4$

$y = 2 x + 2$

$x = - 4$

$y = 2 x + 2$

$x = - 4$

$y = 2 x + 2$

$x = - 4$

$y = 2 x + 2$

Step 3

Replace all occurrences of $x$ with $- 4$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = 2 x + 2$ with $- 4$ .

$y = 2 (- 4) + 2$

$x = - 4$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $2 (- 4) + 2$ .

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Step 3.2.1.1

Multiply $2$ by $- 4$ .

$y = - 8 + 2$

$x = - 4$

Step 3.2.1.2

Add $- 8$ and $2$ .

$y = - 6$

$x = - 4$

$y = - 6$

$x = - 4$

$y = - 6$

$x = - 4$

$y = - 6$

$x = - 4$

Step 4

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

$(- 4, - 6)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 4, - 6)$

Equation Form:

$x = - 4, y = - 6$

Step 6