Algebra Examples

$2 x = y - 10$ $x + 7 = y$

Step 1

Divide each term in $2 x = y - 10$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 x = y - 10$ by $2$ .

$\frac{2 x}{2} = \frac{y}{2} + \frac{- 10}{2}$

$x + 7 = y$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{y}{2} + \frac{- 10}{2}$

$x + 7 = y$

Step 1.2.1.2

Divide $x$ by $1$ .

$x = \frac{y}{2} + \frac{- 10}{2}$

$x + 7 = y$

$x = \frac{y}{2} + \frac{- 10}{2}$

$x + 7 = y$

$x = \frac{y}{2} + \frac{- 10}{2}$

$x + 7 = y$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $- 10$ by $2$ .

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

$x + 7 = y$

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

$x + 7 = y$

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

$x + 7 = y$

Step 2

Replace all occurrences of $x$ with $\frac{y}{2} - 5$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x + 7 = y$ with $\frac{y}{2} - 5$ .

$(\frac{y}{2} - 5) + 7 = y$

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Add $- 5$ and $7$ .

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

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

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

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

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

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

Step 3

Solve for $y$ in $\frac{y}{2} + 2 = y$ .

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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

Subtract $y$ from both sides of the equation.

$\frac{y}{2} + 2 - y = 0$

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

Step 3.1.2

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

$\frac{y}{2} - y \cdot \frac{2}{2} + 2 = 0$

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

Step 3.1.3

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

$\frac{y}{2} + \frac{- y \cdot 2}{2} + 2 = 0$

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

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{y - y \cdot 2}{2} + 2 = 0$

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

Step 3.1.5

Simplify each term.

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Step 3.1.5.1

Simplify the numerator.

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Step 3.1.5.1.1

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

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Step 3.1.5.1.1.1

Raise $y$ to the power of $1$ .

$\frac{y - y \cdot 2}{2} + 2 = 0$

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

Step 3.1.5.1.1.2

Factor $y$ out of $y^{1}$ .

$\frac{y \cdot 1 - y \cdot 2}{2} + 2 = 0$

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

Step 3.1.5.1.1.3

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

$\frac{y \cdot 1 + y (- 1 \cdot 2)}{2} + 2 = 0$

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

Step 3.1.5.1.1.4

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

$\frac{y (1 - 1 \cdot 2)}{2} + 2 = 0$

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

$\frac{y (1 - 1 \cdot 2)}{2} + 2 = 0$

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

Step 3.1.5.1.2

Multiply $- 1$ by $2$ .

$\frac{y (1 - 2)}{2} + 2 = 0$

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

Step 3.1.5.1.3

Subtract $2$ from $1$ .

$\frac{y \cdot - 1}{2} + 2 = 0$

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

$\frac{y \cdot - 1}{2} + 2 = 0$

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

Step 3.1.5.2

Move $- 1$ to the left of $y$ .

$\frac{- 1 \cdot y}{2} + 2 = 0$

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

Step 3.1.5.3

Move the negative in front of the fraction.

$- \frac{y}{2} + 2 = 0$

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

$- \frac{y}{2} + 2 = 0$

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

$- \frac{y}{2} + 2 = 0$

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

Step 3.2

Subtract $2$ from both sides of the equation.

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

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

Step 3.3

Multiply both sides of the equation by $- 2$ .

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

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

Step 3.4

Simplify both sides of the equation.

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Step 3.4.1

Simplify the left side.

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Step 3.4.1.1

Simplify $- 2 (- \frac{y}{2})$ .

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Step 3.4.1.1.1

Cancel the common factor of $2$ .

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Step 3.4.1.1.1.1

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

$- 2 \frac{- y}{2} = - 2 \cdot - 2$

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

Step 3.4.1.1.1.2

Factor $2$ out of $- 2$ .

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

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

Step 3.4.1.1.1.3

Cancel the common factor.

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

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

Step 3.4.1.1.1.4

Rewrite the expression.

$y = - 2 \cdot - 2$

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

$y = - 2 \cdot - 2$

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

Step 3.4.1.1.2

Multiply.

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Step 3.4.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 y = - 2 \cdot - 2$

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

Step 3.4.1.1.2.2

Multiply $y$ by $1$ .

$y = - 2 \cdot - 2$

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

$y = - 2 \cdot - 2$

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

$y = - 2 \cdot - 2$

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

$y = - 2 \cdot - 2$

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

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Multiply $- 2$ by $- 2$ .

$y = 4$

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

$y = 4$

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

$y = 4$

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

$y = 4$

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

Step 4

Replace all occurrences of $y$ with $4$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{y}{2} - 5$ with $4$ .

$x = \frac{4}{2} - 5$

$y = 4$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{4}{2} - 5$ .

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Step 4.2.1.1

Divide $4$ by $2$ .

$x = 2 - 5$

$y = 4$

Step 4.2.1.2

Subtract $5$ from $2$ .

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

Step 5

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

$(- 3, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 3, 4)$

Equation Form:

$x = - 3, y = 4$

Step 7