Algebra Examples

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

Step 1

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

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Step 1.1

Subtract $\frac{1}{y}$ from both sides of the equation.

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

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

Step 1.2

Find the LCD of the terms in the equation.

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Step 1.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x, 1, y$

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

Step 1.2.2

Since $x, 1, y$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part $x^{1}, y^{1}$ .

Since $x, 1, y$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part x,y.

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

Step 1.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

$1$ List the prime factors of each number.

$2$ Multiply each factor the greatest number of times it occurs in either number.

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

Step 1.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 1.2.5

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

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

Step 1.2.6

The factor for $x^{1}$ is $x$ itself.

$x = x$

x occurs $1$ time.

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

Step 1.2.7

The factor for $y^{1}$ is $y$ itself.

$y = y$

y occurs $1$ time.

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

Step 1.2.8

The LCM of $x^{1}, y^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot y$

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

Step 1.2.9

Multiply $x$ by $y$ .

$x y$

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

$x y$

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

Step 1.3

Multiply each term in $\frac{2}{x} = 4 - \frac{1}{y}$ by $x y$ to eliminate the fractions.

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Step 1.3.1

Multiply each term in $\frac{2}{x} = 4 - \frac{1}{y}$ by $x y$ .

$\frac{2}{x} \cdot (x y) = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

Step 1.3.2

Simplify the left side.

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Step 1.3.2.1

Cancel the common factor of $x$ .

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Step 1.3.2.1.1

Factor $x$ out of $x y$ .

$\frac{2}{x} \cdot (x (y)) = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

Step 1.3.2.1.2

Cancel the common factor.

$\frac{2}{x} \cdot (x y) = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

Step 1.3.2.1.3

Rewrite the expression.

$2 y = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

$2 y = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

$2 y = 4 (x y) - \frac{1}{y} \cdot (x y)$

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

Step 1.3.3

Simplify the right side.

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Step 1.3.3.1

Cancel the common factor of $y$ .

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Step 1.3.3.1.1

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

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

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

Step 1.3.3.1.2

Factor $y$ out of $x y$ .

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

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

Step 1.3.3.1.3

Cancel the common factor.

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

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

Step 1.3.3.1.4

Rewrite the expression.

$2 y = 4 x y - x$

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

$2 y = 4 x y - x$

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

$2 y = 4 x y - x$

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

$2 y = 4 x y - x$

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

Step 1.4

Solve the equation.

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Step 1.4.1

Rewrite the equation as $4 x y - x = 2 y$ .

$4 x y - x = 2 y$

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

Step 1.4.2

Factor $x$ out of $4 x y - x$ .

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Step 1.4.2.1

Factor $x$ out of $4 x y$ .

$x (4 y) - x = 2 y$

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

Step 1.4.2.2

Factor $x$ out of $- x$ .

$x (4 y) + x \cdot - 1 = 2 y$

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

Step 1.4.2.3

Factor $x$ out of $x (4 y) + x \cdot - 1$ .

$x (4 y - 1) = 2 y$

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

$x (4 y - 1) = 2 y$

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

Step 1.4.3

Divide each term in $x (4 y - 1) = 2 y$ by $4 y - 1$ and simplify.

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Step 1.4.3.1

Divide each term in $x (4 y - 1) = 2 y$ by $4 y - 1$ .

$\frac{x (4 y - 1)}{4 y - 1} = \frac{2 y}{4 y - 1}$

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

Step 1.4.3.2

Simplify the left side.

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Step 1.4.3.2.1

Cancel the common factor of $4 y - 1$ .

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Step 1.4.3.2.1.1

Cancel the common factor.

$\frac{x (4 y - 1)}{4 y - 1} = \frac{2 y}{4 y - 1}$

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

Step 1.4.3.2.1.2

Divide $x$ by $1$ .

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

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

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

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

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

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

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

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

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

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

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

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

Step 2

Replace all occurrences of $x$ with $\frac{2 y}{4 y - 1}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $\frac{3}{x} + \frac{5}{y} = - 1$ with $\frac{2 y}{4 y - 1}$ .

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $\frac{3}{\frac{2 y}{4 y - 1}} + \frac{5}{y}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Multiply the numerator by the reciprocal of the denominator.

$3 (\frac{4 y - 1}{2 y}) + \frac{5}{y} = - 1$

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

Step 2.2.1.1.2

Combine $3$ and $\frac{4 y - 1}{2 y}$ .

$\frac{3 (4 y - 1)}{2 y} + \frac{5}{y} = - 1$

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

$\frac{3 (4 y - 1)}{2 y} + \frac{5}{y} = - 1$

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

Step 2.2.1.2

To write $\frac{5}{y}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3 (4 y - 1)}{2 y} + \frac{5}{y} \cdot \frac{2}{2} = - 1$

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

Step 2.2.1.3

Write each expression with a common denominator of $2 y$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.2.1.3.1

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

$\frac{3 (4 y - 1)}{2 y} + \frac{5 \cdot 2}{y \cdot 2} = - 1$

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

Step 2.2.1.3.2

Reorder the factors of $y \cdot 2$ .

$\frac{3 (4 y - 1)}{2 y} + \frac{5 \cdot 2}{2 y} = - 1$

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

$\frac{3 (4 y - 1)}{2 y} + \frac{5 \cdot 2}{2 y} = - 1$

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

Step 2.2.1.4

Combine the numerators over the common denominator.

$\frac{3 (4 y - 1) + 5 \cdot 2}{2 y} = - 1$

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

Step 2.2.1.5

Simplify the numerator.

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Step 2.2.1.5.1

Apply the distributive property.

$\frac{3 (4 y) + 3 \cdot - 1 + 5 \cdot 2}{2 y} = - 1$

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

Step 2.2.1.5.2

Multiply $4$ by $3$ .

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

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

Step 2.2.1.5.3

Multiply $3$ by $- 1$ .

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

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

Step 2.2.1.5.4

Multiply $5$ by $2$ .

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

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

Step 2.2.1.5.5

Add $- 3$ and $10$ .

$\frac{12 y + 7}{2 y} = - 1$

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

$\frac{12 y + 7}{2 y} = - 1$

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

$\frac{12 y + 7}{2 y} = - 1$

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

$\frac{12 y + 7}{2 y} = - 1$

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

$\frac{12 y + 7}{2 y} = - 1$

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

Step 3

Solve for $y$ in $\frac{12 y + 7}{2 y} = - 1$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2 y, 1$

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

Step 3.1.2

The LCM of one and any expression is the expression.

$2 y$

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

$2 y$

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

Step 3.2

Multiply each term in $\frac{12 y + 7}{2 y} = - 1$ by $2 y$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{12 y + 7}{2 y} = - 1$ by $2 y$ .

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

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Rewrite using the commutative property of multiplication.

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

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

Step 3.2.2.2

Cancel the common factor of $2$ .

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Step 3.2.2.2.1

Factor $2$ out of $2 y$ .

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

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

Step 3.2.2.2.2

Cancel the common factor.

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

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

Step 3.2.2.2.3

Rewrite the expression.

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

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

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

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

Step 3.2.2.3

Cancel the common factor of $y$ .

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Step 3.2.2.3.1

Cancel the common factor.

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

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

Step 3.2.2.3.2

Rewrite the expression.

$12 y + 7 = - (2 y)$

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

$12 y + 7 = - (2 y)$

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

$12 y + 7 = - (2 y)$

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

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Multiply $2$ by $- 1$ .

$12 y + 7 = - 2 y$

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

$12 y + 7 = - 2 y$

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

$12 y + 7 = - 2 y$

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

Step 3.3

Solve the equation.

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Step 3.3.1

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

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Step 3.3.1.1

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

$12 y + 7 + 2 y = 0$

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

Step 3.3.1.2

Add $12 y$ and $2 y$ .

$14 y + 7 = 0$

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

$14 y + 7 = 0$

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

Step 3.3.2

Subtract $7$ from both sides of the equation.

$14 y = - 7$

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

Step 3.3.3

Divide each term in $14 y = - 7$ by $14$ and simplify.

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Step 3.3.3.1

Divide each term in $14 y = - 7$ by $14$ .

$\frac{14 y}{14} = \frac{- 7}{14}$

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

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $14$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{14 y}{14} = \frac{- 7}{14}$

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

Step 3.3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 7}{14}$

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

$y = \frac{- 7}{14}$

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

$y = \frac{- 7}{14}$

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

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Cancel the common factor of $- 7$ and $14$ .

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Step 3.3.3.3.1.1

Factor $7$ out of $- 7$ .

$y = \frac{7 (- 1)}{14}$

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

Step 3.3.3.3.1.2

Cancel the common factors.

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Step 3.3.3.3.1.2.1

Factor $7$ out of $14$ .

$y = \frac{7 \cdot - 1}{7 \cdot 2}$

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

Step 3.3.3.3.1.2.2

Cancel the common factor.

$y = \frac{7 \cdot - 1}{7 \cdot 2}$

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

Step 3.3.3.3.1.2.3

Rewrite the expression.

$y = \frac{- 1}{2}$

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

$y = \frac{- 1}{2}$

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

$y = \frac{- 1}{2}$

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

Step 3.3.3.3.2

Move the negative in front of the fraction.

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

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

Step 4

Replace all occurrences of $y$ with $- \frac{1}{2}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{2 y}{4 y - 1}$ with $- \frac{1}{2}$ .

$x = \frac{2 (- \frac{1}{2})}{4 (- \frac{1}{2}) - 1}$

$y = - \frac{1}{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{2 (- \frac{1}{2})}{4 (- \frac{1}{2}) - 1}$ .

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Step 4.2.1.1

Simplify the numerator.

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Step 4.2.1.1.1

Multiply $- 1$ by $2$ .

$x = \frac{- 2 (\frac{1}{2})}{4 (- \frac{1}{2}) - 1}$

$y = - \frac{1}{2}$

Step 4.2.1.1.2

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

$x = \frac{\frac{- 2}{2}}{4 (- \frac{1}{2}) - 1}$

$y = - \frac{1}{2}$

$x = \frac{\frac{- 2}{2}}{4 (- \frac{1}{2}) - 1}$

$y = - \frac{1}{2}$

Step 4.2.1.2

Simplify the denominator.

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Step 4.2.1.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.2.1.1

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

$x = \frac{\frac{- 2}{2}}{4 (\frac{- 1}{2}) - 1}$

$y = - \frac{1}{2}$

Step 4.2.1.2.1.2

Factor $2$ out of $4$ .

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

$y = - \frac{1}{2}$

Step 4.2.1.2.1.3

Cancel the common factor.

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

$y = - \frac{1}{2}$

Step 4.2.1.2.1.4

Rewrite the expression.

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

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

Step 4.2.1.2.2

Multiply $2$ by $- 1$ .

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

$y = - \frac{1}{2}$

Step 4.2.1.2.3

Subtract $1$ from $- 2$ .

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

$y = - \frac{1}{2}$

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

$y = - \frac{1}{2}$

Step 4.2.1.3

Reduce the expression by cancelling the common factors.

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Step 4.2.1.3.1

Divide $- 2$ by $2$ .

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

$y = - \frac{1}{2}$

Step 4.2.1.3.2

Dividing two negative values results in a positive value.

$x = \frac{1}{3}$

$y = - \frac{1}{2}$

$x = \frac{1}{3}$

$y = - \frac{1}{2}$

$x = \frac{1}{3}$

$y = - \frac{1}{2}$

$x = \frac{1}{3}$

$y = - \frac{1}{2}$

$x = \frac{1}{3}$

$y = - \frac{1}{2}$

Step 5

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

$(\frac{1}{3}, - \frac{1}{2})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{3}, - \frac{1}{2})$

Equation Form:

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

Step 7