Solve the System of Equations xy=-4 2y=x-6

$x y = - 4$ $2 y = x - 6$

Divide each term in $x y = - 4$ by $y$ and simplify.

Divide each term in $x y = - 4$ by $y$ .

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

$2 y = x - 6$

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

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

$2 y = x - 6$

Divide $x$ by $1$ .

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

$2 y = x - 6$

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

$2 y = x - 6$

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

$2 y = x - 6$

Simplify the right side.

Move the negative in front of the fraction.

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

$2 y = x - 6$

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

$2 y = x - 6$

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

$2 y = x - 6$

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

Replace all occurrences of $x$ in $2 y = x - 6$ with $- \frac{4}{y}$ .

$2 y = (- \frac{4}{y}) - 6$

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

Simplify the right side.

Remove parentheses.

$2 y = - \frac{4}{y} - 6$

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

$2 y = - \frac{4}{y} - 6$

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

$2 y = - \frac{4}{y} - 6$

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

Solve for $y$ in $2 y = - \frac{4}{y} - 6$ .

Find the LCD of the terms in the equation.

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

$1, y, 1$

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

The LCM of one and any expression is the expression.

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

Multiply each term in $2 y = - \frac{4}{y} - 6$ by $y$ to eliminate the fractions.

Multiply each term in $2 y = - \frac{4}{y} - 6$ by $y$ .

$2 y \cdot y = - \frac{4}{y} \cdot y - 6 y$

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

Simplify the left side.

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$2 (y \cdot y) = - \frac{4}{y} \cdot y - 6 y$

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

Multiply $y$ by $y$ .

$2 y^{2} = - \frac{4}{y} \cdot y - 6 y$

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

$2 y^{2} = - \frac{4}{y} \cdot y - 6 y$

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

$2 y^{2} = - \frac{4}{y} \cdot y - 6 y$

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

Simplify the right side.

Cancel the common factor of $y$ .

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

$2 y^{2} = \frac{- 4}{y} \cdot y - 6 y$

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

Cancel the common factor.

$2 y^{2} = \frac{- 4}{y} \cdot y - 6 y$

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

Rewrite the expression.

$2 y^{2} = - 4 - 6 y$

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

$2 y^{2} = - 4 - 6 y$

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

$2 y^{2} = - 4 - 6 y$

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

$2 y^{2} = - 4 - 6 y$

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

Solve the equation.

Add $6 y$ to both sides of the equation.

$2 y^{2} + 6 y = - 4$

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

Add $4$ to both sides of the equation.

$2 y^{2} + 6 y + 4 = 0$

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

Factor the left side of the equation.

Factor $2$ out of $2 y^{2} + 6 y + 4$ .

Factor $2$ out of $2 y^{2}$ .

$2 (y^{2}) + 6 y + 4 = 0$

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

Factor $2$ out of $6 y$ .

$2 (y^{2}) + 2 (3 y) + 4 = 0$

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

Factor $2$ out of $4$ .

$2 y^{2} + 2 (3 y) + 2 \cdot 2 = 0$

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

Factor $2$ out of $2 y^{2} + 2 (3 y)$ .

$2 (y^{2} + 3 y) + 2 \cdot 2 = 0$

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

Factor $2$ out of $2 (y^{2} + 3 y) + 2 \cdot 2$ .

$2 (y^{2} + 3 y + 2) = 0$

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

$2 (y^{2} + 3 y + 2) = 0$

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

Factor.

Factor $y^{2} + 3 y + 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

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

Write the factored form using these integers.

$2 ((y + 1) (y + 2)) = 0$

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

$2 ((y + 1) (y + 2)) = 0$

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

Remove unnecessary parentheses.

$2 (y + 1) (y + 2) = 0$

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

$2 (y + 1) (y + 2) = 0$

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

$2 (y + 1) (y + 2) = 0$

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y + 1 = 0$

$y + 2 = 0$

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

Set $y + 1$ equal to $0$ and solve for $y$ .

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$y = - 1$

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

$y = - 1$

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

Set $y + 2$ equal to $0$ and solve for $y$ .

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$y = - 2$

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

$y = - 2$

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

The final solution is all the values that make $2 (y + 1) (y + 2) = 0$ true.

$y = - 1, - 2$

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

$y = - 1, - 2$

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

$y = - 1, - 2$

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

Replace all occurrences of $y$ with $- 1$ in each equation.

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

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

$y = - 1$

Simplify the right side.

Simplify $- \frac{4}{- 1}$ .

Divide $4$ by $- 1$ .

$x = 4$

$y = - 1$

Multiply $- 1$ by $- 4$ .

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

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

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

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

$y = - 2$

Simplify the right side.

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

Divide $4$ by $- 2$ .

$x = 2$

$y = - 2$

Multiply $- 1$ by $- 2$ .

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

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

$(4, - 1)$

$(2, - 2)$

The result can be shown in multiple forms.

Point Form:

$(4, - 1), (2, - 2)$

Equation Form:

$x = 4, y = - 1$

$x = 2, y = - 2$

Solve the System of Equations xy=-4 2y=x-6

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