Algebra Examples

$\frac{1}{3} x - \frac{1}{2} y = - 12$ $2 x - \frac{1}{5} y = \frac{4}{5}$

Step 1

Solve for $x$ in $2 x - \frac{1}{5} y = \frac{4}{5}$ .

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Step 1.1

Combine $y$ and $\frac{1}{5}$ .

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.2

Add $\frac{y}{5}$ to both sides of the equation.

$2 x = \frac{4}{5} + \frac{y}{5}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3

Divide each term in $2 x = \frac{4}{5} + \frac{y}{5}$ by $2$ and simplify.

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Step 1.3.1

Divide each term in $2 x = \frac{4}{5} + \frac{y}{5}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{4}{5}}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{4}{5}}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{4}{5}}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{\frac{4}{5}}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{\frac{4}{5}}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3

Simplify the right side.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{4}{5} \cdot \frac{1}{2} + \frac{\frac{y}{5}}{2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.2

Cancel the common factor of $2$ .

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Step 1.3.3.1.2.1

Factor $2$ out of $4$ .

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.2.2

Cancel the common factor.

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.2.3

Rewrite the expression.

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.3

Multiply the numerator by the reciprocal of the denominator.

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

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.4

Multiply $\frac{y}{5} \cdot \frac{1}{2}$ .

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Step 1.3.3.1.4.1

Multiply $\frac{y}{5}$ by $\frac{1}{2}$ .

$x = \frac{2}{5} + \frac{y}{5 \cdot 2}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 1.3.3.1.4.2

Multiply $5$ by $2$ .

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{1}{3} x - \frac{1}{2} y = - 12$

Step 2

Replace all occurrences of $x$ with $\frac{2}{5} + \frac{y}{10}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $\frac{1}{3} x - \frac{1}{2} y = - 12$ with $\frac{2}{5} + \frac{y}{10}$ .

$\frac{1}{3} \cdot (\frac{2}{5} + \frac{y}{10}) - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $\frac{1}{3} \cdot (\frac{2}{5} + \frac{y}{10}) - \frac{1}{2} y$ .

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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.

$\frac{1}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{y}{10} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.1.2

Multiply $\frac{1}{3} \cdot \frac{2}{5}$ .

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Step 2.2.1.1.2.1

Multiply $\frac{1}{3}$ by $\frac{2}{5}$ .

$\frac{2}{3 \cdot 5} + \frac{1}{3} \cdot \frac{y}{10} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.1.2.2

Multiply $3$ by $5$ .

$\frac{2}{15} + \frac{1}{3} \cdot \frac{y}{10} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{1}{3} \cdot \frac{y}{10} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.1.3

Multiply $\frac{1}{3} \cdot \frac{y}{10}$ .

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Step 2.2.1.1.3.1

Multiply $\frac{1}{3}$ by $\frac{y}{10}$ .

$\frac{2}{15} + \frac{y}{3 \cdot 10} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.1.3.2

Multiply $3$ by $10$ .

$\frac{2}{15} + \frac{y}{30} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y}{30} - \frac{1}{2} y = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.1.4

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

$\frac{2}{15} + \frac{y}{30} - \frac{y}{2} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y}{30} - \frac{y}{2} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.2

To write $- \frac{y}{2}$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$\frac{2}{15} + \frac{y}{30} - \frac{y}{2} \cdot \frac{15}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.3

Write each expression with a common denominator of $30$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.2.1.3.1

Multiply $\frac{y}{2}$ by $\frac{15}{15}$ .

$\frac{2}{15} + \frac{y}{30} - \frac{y \cdot 15}{2 \cdot 15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.3.2

Multiply $2$ by $15$ .

$\frac{2}{15} + \frac{y}{30} - \frac{y \cdot 15}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y}{30} - \frac{y \cdot 15}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.4

Combine the numerators over the common denominator.

$\frac{2}{15} + \frac{y - y \cdot 15}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5

Simplify each term.

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Step 2.2.1.5.1

Simplify the numerator.

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Step 2.2.1.5.1.1

Factor $y$ out of $y - y \cdot 15$ .

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Step 2.2.1.5.1.1.1

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

$\frac{2}{15} + \frac{y - y \cdot 15}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.1.1.2

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

$\frac{2}{15} + \frac{y \cdot 1 - y \cdot 15}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.1.1.3

Factor $y$ out of $- y \cdot 15$ .

$\frac{2}{15} + \frac{y \cdot 1 + y (- 1 \cdot 15)}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.1.1.4

Factor $y$ out of $y \cdot 1 + y (- 1 \cdot 15)$ .

$\frac{2}{15} + \frac{y (1 - 1 \cdot 15)}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y (1 - 1 \cdot 15)}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.1.2

Multiply $- 1$ by $15$ .

$\frac{2}{15} + \frac{y (1 - 15)}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.1.3

Subtract $15$ from $1$ .

$\frac{2}{15} + \frac{y \cdot - 14}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y \cdot - 14}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.2

Cancel the common factor of $- 14$ and $30$ .

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Step 2.2.1.5.2.1

Factor $2$ out of $y \cdot - 14$ .

$\frac{2}{15} + \frac{2 (y \cdot - 7)}{30} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.2.2

Cancel the common factors.

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Step 2.2.1.5.2.2.1

Factor $2$ out of $30$ .

$\frac{2}{15} + \frac{2 (y \cdot - 7)}{2 (15)} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.2.2.2

Cancel the common factor.

$\frac{2}{15} + \frac{2 (y \cdot - 7)}{2 \cdot 15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.2.2.3

Rewrite the expression.

$\frac{2}{15} + \frac{y \cdot - 7}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y \cdot - 7}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} + \frac{y \cdot - 7}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.3

Move $- 7$ to the left of $y$ .

$\frac{2}{15} + \frac{- 7 \cdot y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 2.2.1.5.4

Move the negative in front of the fraction.

$\frac{2}{15} - \frac{7 y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} - \frac{7 y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} - \frac{7 y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} - \frac{7 y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

$\frac{2}{15} - \frac{7 y}{15} = - 12$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3

Solve for $y$ in $\frac{2}{15} - \frac{7 y}{15} = - 12$ .

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Step 3.1

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

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Step 3.1.1

Subtract $\frac{2}{15}$ from both sides of the equation.

$- \frac{7 y}{15} = - 12 - \frac{2}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.2

To write $- 12$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$- \frac{7 y}{15} = - 12 \cdot \frac{15}{15} - \frac{2}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.3

Combine $- 12$ and $\frac{15}{15}$ .

$- \frac{7 y}{15} = \frac{- 12 \cdot 15}{15} - \frac{2}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.4

Combine the numerators over the common denominator.

$- \frac{7 y}{15} = \frac{- 12 \cdot 15 - 2}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Multiply $- 12$ by $15$ .

$- \frac{7 y}{15} = \frac{- 180 - 2}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.5.2

Subtract $2$ from $- 180$ .

$- \frac{7 y}{15} = \frac{- 182}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

$- \frac{7 y}{15} = \frac{- 182}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.1.6

Move the negative in front of the fraction.

$- \frac{7 y}{15} = - \frac{182}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

$- \frac{7 y}{15} = - \frac{182}{15}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- 7 y = - 182$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.3

Divide each term in $- 7 y = - 182$ by $- 7$ and simplify.

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Step 3.3.1

Divide each term in $- 7 y = - 182$ by $- 7$ .

$\frac{- 7 y}{- 7} = \frac{- 182}{- 7}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 7$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 7 y}{- 7} = \frac{- 182}{- 7}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 182}{- 7}$

$x = \frac{2}{5} + \frac{y}{10}$

$y = \frac{- 182}{- 7}$

$x = \frac{2}{5} + \frac{y}{10}$

$y = \frac{- 182}{- 7}$

$x = \frac{2}{5} + \frac{y}{10}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $- 182$ by $- 7$ .

$y = 26$

$x = \frac{2}{5} + \frac{y}{10}$

$y = 26$

$x = \frac{2}{5} + \frac{y}{10}$

$y = 26$

$x = \frac{2}{5} + \frac{y}{10}$

$y = 26$

$x = \frac{2}{5} + \frac{y}{10}$

Step 4

Replace all occurrences of $y$ with $26$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{2}{5} + \frac{y}{10}$ with $26$ .

$x = \frac{2}{5} + \frac{26}{10}$

$y = 26$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{2}{5} + \frac{26}{10}$ .

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Step 4.2.1.1

Cancel the common factor of $26$ and $10$ .

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Step 4.2.1.1.1

Factor $2$ out of $26$ .

$x = \frac{2}{5} + \frac{2 (13)}{10}$

$y = 26$

Step 4.2.1.1.2

Cancel the common factors.

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Step 4.2.1.1.2.1

Factor $2$ out of $10$ .

$x = \frac{2}{5} + \frac{2 \cdot 13}{2 \cdot 5}$

$y = 26$

Step 4.2.1.1.2.2

Cancel the common factor.

$x = \frac{2}{5} + \frac{2 \cdot 13}{2 \cdot 5}$

$y = 26$

Step 4.2.1.1.2.3

Rewrite the expression.

$x = \frac{2}{5} + \frac{13}{5}$

$y = 26$

$x = \frac{2}{5} + \frac{13}{5}$

$y = 26$

$x = \frac{2}{5} + \frac{13}{5}$

$y = 26$

Step 4.2.1.2

Combine the numerators over the common denominator.

$x = \frac{2 + 13}{5}$

$y = 26$

Step 4.2.1.3

Simplify the expression.

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Step 4.2.1.3.1

Add $2$ and $13$ .

$x = \frac{15}{5}$

$y = 26$

Step 4.2.1.3.2

Divide $15$ by $5$ .

$x = 3$

$y = 26$

$x = 3$

$y = 26$

$x = 3$

$y = 26$

$x = 3$

$y = 26$

$x = 3$

$y = 26$

Step 5

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

$(3, 26)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 26)$

Equation Form:

$x = 3, y = 26$

Step 7