Precalculus Examples

$x^{2} + 2 x y + y^{2} = 36$ , $x^{2} - x y = 0$

Step 1

Subtract $36$ from both sides of the equation.

$x^{2} + 2 x y + y^{2} - 36 = 0$

$x^{2} - x y = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

$x^{2} - x y = 0$

Step 3

Substitute the values $a = 1$ , $b = 2 x$ , and $c = x^{2} - 36$ into the quadratic formula and solve for $y$ .

$\frac{- 2 x \pm \sqrt{{(2 x)}^{2} - 4 \cdot (1 \cdot (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Step 4

Simplify.

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Simplify the numerator.

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Add parentheses.

$y = \frac{- 2 x \pm \sqrt{{(2 x)}^{2} - 4 \cdot (1 \cdot (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Let $u = 1 \cdot (x^{2} - 36)$ . Substitute $u$ for all occurrences of $1 \cdot (x^{2} - 36)$ .

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Apply the product rule to $2 x$ .

$y = \frac{- 2 x \pm \sqrt{2^{2} x^{2} - 4 \cdot u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Raise $2$ to the power of $2$ .

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $4 x^{2} - 4 u$ .

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Factor $4$ out of $4 x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2}) - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $- 4 u$ .

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

$x^{2} - x y = 0$

Factor $4$ out of $4 (x^{2}) + 4 (- u)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Replace all occurrences of $u$ with $1 \cdot (x^{2} - 36)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (1 \cdot (x^{2} - 36)))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Simplify.

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Simplify each term.

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Multiply $x^{2} - 36$ by $1$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Apply the distributive property.

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $- 1$ by $- 36$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Subtract $x^{2}$ from $x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (0 + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Add $0$ and $36$ .

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $4$ by $36$ .

$y = \frac{- 2 x \pm \sqrt{144}}{2 \cdot 1}$

$x^{2} - x y = 0$

Rewrite $144$ as $12^{2}$ .

$y = \frac{- 2 x \pm \sqrt{12^{2}}}{2 \cdot 1}$

$x^{2} - x y = 0$

Pull terms out from under the radical, assuming positive real numbers.

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

$x^{2} - x y = 0$

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

$x^{2} - x y = 0$

Multiply $2$ by $1$ .

$y = \frac{- 2 x \pm 12}{2}$

$x^{2} - x y = 0$

Simplify $\frac{- 2 x \pm 12}{2}$ .

$y = - x \pm 6$

$x^{2} - x y = 0$

$y = - x \pm 6$

$x^{2} - x y = 0$

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Add parentheses.

$y = \frac{- 2 x \pm \sqrt{{(2 x)}^{2} - 4 \cdot (1 \cdot (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Let $u = 1 \cdot (x^{2} - 36)$ . Substitute $u$ for all occurrences of $1 \cdot (x^{2} - 36)$ .

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Apply the product rule to $2 x$ .

$y = \frac{- 2 x \pm \sqrt{2^{2} x^{2} - 4 \cdot u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Raise $2$ to the power of $2$ .

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $4 x^{2} - 4 u$ .

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Factor $4$ out of $4 x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2}) - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $- 4 u$ .

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

$x^{2} - x y = 0$

Factor $4$ out of $4 (x^{2}) + 4 (- u)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Replace all occurrences of $u$ with $1 \cdot (x^{2} - 36)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (1 \cdot (x^{2} - 36)))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Simplify.

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Simplify each term.

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Multiply $x^{2} - 36$ by $1$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Apply the distributive property.

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $- 1$ by $- 36$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Subtract $x^{2}$ from $x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (0 + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Add $0$ and $36$ .

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $4$ by $36$ .

$y = \frac{- 2 x \pm \sqrt{144}}{2 \cdot 1}$

$x^{2} - x y = 0$

Rewrite $144$ as $12^{2}$ .

$y = \frac{- 2 x \pm \sqrt{12^{2}}}{2 \cdot 1}$

$x^{2} - x y = 0$

Pull terms out from under the radical, assuming positive real numbers.

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

$x^{2} - x y = 0$

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

$x^{2} - x y = 0$

Multiply $2$ by $1$ .

$y = \frac{- 2 x \pm 12}{2}$

$x^{2} - x y = 0$

Simplify $\frac{- 2 x \pm 12}{2}$ .

$y = - x \pm 6$

$x^{2} - x y = 0$

Change the $\pm$ to $+$ .

$y = - x + 6$

$x^{2} - x y = 0$

$y = - x + 6$

$x^{2} - x y = 0$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Add parentheses.

$y = \frac{- 2 x \pm \sqrt{{(2 x)}^{2} - 4 \cdot (1 \cdot (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Let $u = 1 \cdot (x^{2} - 36)$ . Substitute $u$ for all occurrences of $1 \cdot (x^{2} - 36)$ .

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Apply the product rule to $2 x$ .

$y = \frac{- 2 x \pm \sqrt{2^{2} x^{2} - 4 \cdot u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Raise $2$ to the power of $2$ .

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 x^{2} - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $4 x^{2} - 4 u$ .

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Factor $4$ out of $4 x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2}) - 4 u}}{2 \cdot 1}$

$x^{2} - x y = 0$

Factor $4$ out of $- 4 u$ .

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

$x^{2} - x y = 0$

Factor $4$ out of $4 (x^{2}) + 4 (- u)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - u)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Replace all occurrences of $u$ with $1 \cdot (x^{2} - 36)$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (1 \cdot (x^{2} - 36)))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Simplify.

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Simplify each term.

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Multiply $x^{2} - 36$ by $1$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - (x^{2} - 36))}}{2 \cdot 1}$

$x^{2} - x y = 0$

Apply the distributive property.

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $- 1$ by $- 36$ .

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 (x^{2} - x^{2} + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Subtract $x^{2}$ from $x^{2}$ .

$y = \frac{- 2 x \pm \sqrt{4 (0 + 36)}}{2 \cdot 1}$

$x^{2} - x y = 0$

Add $0$ and $36$ .

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

$y = \frac{- 2 x \pm \sqrt{4 \cdot 36}}{2 \cdot 1}$

$x^{2} - x y = 0$

Multiply $4$ by $36$ .

$y = \frac{- 2 x \pm \sqrt{144}}{2 \cdot 1}$

$x^{2} - x y = 0$

Rewrite $144$ as $12^{2}$ .

$y = \frac{- 2 x \pm \sqrt{12^{2}}}{2 \cdot 1}$

$x^{2} - x y = 0$

Pull terms out from under the radical, assuming positive real numbers.

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

$x^{2} - x y = 0$

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

$x^{2} - x y = 0$

Multiply $2$ by $1$ .

$y = \frac{- 2 x \pm 12}{2}$

$x^{2} - x y = 0$

Simplify $\frac{- 2 x \pm 12}{2}$ .

$y = - x \pm 6$

$x^{2} - x y = 0$

Change the $\pm$ to $-$ .

$y = - x - 6$

$x^{2} - x y = 0$

$y = - x - 6$

$x^{2} - x y = 0$

Step 7

The final answer is the combination of both solutions.

$y = - x + 6$

$y = - x - 6$

$x^{2} - x y = 0$

Step 8

Subtract $x^{2}$ from both sides of the equation.

$- x y = - x^{2}$

$y = - x + 6$

$y = - x - 6$

Step 9

Divide each term in $- x y = - x^{2}$ by $- 1 x$ and simplify.

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Divide each term in $- x y = - x^{2}$ by $- 1 x$ .

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

$y = - x + 6$

$y = - x - 6$

Simplify the left side.

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Dividing two negative values results in a positive value.

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

$y = - x + 6$

$y = - x - 6$

Cancel the common factor of $x$ .

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Cancel the common factor.

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

$y = - x + 6$

$y = - x - 6$

Divide $y$ by $1$ .

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

$y = - x + 6$

$y = - x - 6$

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

$y = - x + 6$

$y = - x - 6$

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

$y = - x + 6$

$y = - x - 6$

Simplify the right side.

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Dividing two negative values results in a positive value.

$y = \frac{x^{2}}{x}$

$y = - x + 6$

$y = - x - 6$

Cancel the common factor of $x^{2}$ and $x$ .

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

$y = \frac{x \cdot x}{x}$

$y = - x + 6$

$y = - x - 6$

Cancel the common factors.

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

$y = \frac{x \cdot x}{x}$

$y = - x + 6$

$y = - x - 6$

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

$y = \frac{x \cdot x}{x \cdot 1}$

$y = - x + 6$

$y = - x - 6$

Cancel the common factor.

$y = \frac{x \cdot x}{x \cdot 1}$

$y = - x + 6$

$y = - x - 6$

Rewrite the expression.

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

$y = - x + 6$

$y = - x - 6$

Divide $x$ by $1$ .

$y = x$

$y = - x + 6$

$y = - x - 6$

$y = x$

$y = - x + 6$

$y = - x - 6$

$y = x$

$y = - x + 6$

$y = - x - 6$

$y = x$

$y = - x + 6$

$y = - x - 6$

$y = x$

$y = - x + 6$

$y = - x - 6$

Step 10

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

$(3, 3)$

$(- 3, - 3)$

Step 11