Algebra Examples

$- \frac{1}{3} x + \frac{1}{2} y = 14$ $2 x + \frac{1}{5} y = - \frac{4}{5}$

Step 1

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

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Step 1.1

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

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.2

Subtract $\frac{y}{5}$ from both sides of the equation.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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 $2$ .

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.2.1.2

Divide $x$ by $1$ .

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.2

Cancel the common factor of $2$ .

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Step 1.3.3.1.2.1

Move the leading negative in $- \frac{4}{5}$ into the numerator.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.2.2

Factor $2$ out of $- 4$ .

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.2.3

Cancel the common factor.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.2.4

Rewrite the expression.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.3

Move the negative in front of the fraction.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.4

Multiply the numerator by the reciprocal of the denominator.

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.5

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

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Step 1.3.3.1.5.1

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

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

Step 1.3.3.1.5.2

Multiply $2$ by $5$ .

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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

$- \frac{1}{3} x + \frac{1}{2} y = 14$

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 $- \frac{1}{3} x + \frac{1}{2} y = 14$ with $- \frac{2}{5} - \frac{y}{10}$ .

$- \frac{1}{3} \cdot (- \frac{2}{5} - \frac{y}{10}) + \frac{1}{2} y = 14$

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

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.

$- \frac{1}{3} \cdot (- \frac{2}{5}) - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

Step 2.2.1.1.2

Multiply $- \frac{1}{3} (- \frac{2}{5})$ .

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Step 2.2.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{3}) \cdot \frac{2}{5} - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

Step 2.2.1.1.2.2

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

$\frac{1}{3} \cdot \frac{2}{5} - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

Step 2.2.1.1.2.3

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

$\frac{2}{3 \cdot 5} - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

Step 2.2.1.1.2.4

Multiply $3$ by $5$ .

$\frac{2}{15} - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

$\frac{2}{15} - \frac{1}{3} \cdot (- \frac{y}{10}) + \frac{1}{2} y = 14$

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

Step 2.2.1.1.3

Multiply $- \frac{1}{3} (- \frac{y}{10})$ .

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Step 2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$\frac{2}{15} + 1 (\frac{1}{3}) \cdot \frac{y}{10} + \frac{1}{2} y = 14$

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

Step 2.2.1.1.3.2

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

$\frac{2}{15} + \frac{1}{3} \cdot \frac{y}{10} + \frac{1}{2} y = 14$

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

Step 2.2.1.1.3.3

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

$\frac{2}{15} + \frac{y}{3 \cdot 10} + \frac{1}{2} y = 14$

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

Step 2.2.1.1.3.4

Multiply $3$ by $10$ .

$\frac{2}{15} + \frac{y}{30} + \frac{1}{2} y = 14$

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

$\frac{2}{15} + \frac{y}{30} + \frac{1}{2} y = 14$

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

Step 2.2.1.1.4

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

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

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

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

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

Step 2.2.1.2

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

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

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

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 $\frac{15}{15}$ .

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

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

Step 2.2.1.3.2

Multiply $2$ by $15$ .

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

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

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

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

Step 2.2.1.4

Combine the numerators over the common denominator.

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

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

Step 2.2.1.5

Add $y$ and $y \cdot 15$ .

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Step 2.2.1.5.1

Reorder $y$ and $15$ .

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

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

Step 2.2.1.5.2

Add $y$ and $15 \cdot y$ .

$\frac{2}{15} + \frac{16 y}{30} = 14$

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

$\frac{2}{15} + \frac{16 y}{30} = 14$

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

Step 2.2.1.6

Cancel the common factor of $16$ and $30$ .

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Step 2.2.1.6.1

Factor $2$ out of $16 y$ .

$\frac{2}{15} + \frac{2 (8 y)}{30} = 14$

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

Step 2.2.1.6.2

Cancel the common factors.

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Step 2.2.1.6.2.1

Factor $2$ out of $30$ .

$\frac{2}{15} + \frac{2 (8 y)}{2 (15)} = 14$

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

Step 2.2.1.6.2.2

Cancel the common factor.

$\frac{2}{15} + \frac{2 (8 y)}{2 \cdot 15} = 14$

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

Step 2.2.1.6.2.3

Rewrite the expression.

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

$\frac{2}{15} + \frac{8 y}{15} = 14$

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

Step 3

Solve for $y$ in $\frac{2}{15} + \frac{8 y}{15} = 14$ .

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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.

$\frac{8 y}{15} = 14 - \frac{2}{15}$

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

Step 3.1.2

To write $14$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$\frac{8 y}{15} = 14 \cdot \frac{15}{15} - \frac{2}{15}$

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

Step 3.1.3

Combine $14$ and $\frac{15}{15}$ .

$\frac{8 y}{15} = \frac{14 \cdot 15}{15} - \frac{2}{15}$

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

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{8 y}{15} = \frac{14 \cdot 15 - 2}{15}$

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

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Multiply $14$ by $15$ .

$\frac{8 y}{15} = \frac{210 - 2}{15}$

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

Step 3.1.5.2

Subtract $2$ from $210$ .

$\frac{8 y}{15} = \frac{208}{15}$

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

$\frac{8 y}{15} = \frac{208}{15}$

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

$\frac{8 y}{15} = \frac{208}{15}$

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

Step 3.2

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

$8 y = 208$

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

Step 3.3

Divide each term in $8 y = 208$ by $8$ and simplify.

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Step 3.3.1

Divide each term in $8 y = 208$ by $8$ .

$\frac{8 y}{8} = \frac{208}{8}$

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $8$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{8 y}{8} = \frac{208}{8}$

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

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{208}{8}$

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

$y = \frac{208}{8}$

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

$y = \frac{208}{8}$

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $208$ by $8$ .

$y = 26$

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

$y = 26$

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

$y = 26$

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

$y = 26$

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

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

Subtract $13$ from $- 2$ .

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

$y = 26$

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