Algebra Examples

$12 x^{2} - 12 y^{2} + x - y - 3 = 0$

Step 1

Add $3$ to both sides of the equation.

$12 x^{2} - 12 y^{2} + x - y = 3$

Step 2

Complete the square for $12 x^{2} + x$ .

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Step 2.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 12$

$b = 1$

$c = 0$

Step 2.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 2.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 2.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 \cdot 12}$

Step 2.3.2

Multiply $2$ by $12$ .

$d = \frac{1}{24}$

Step 2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{1^{2}}{4 \cdot 12}$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

One to any power is one.

$e = 0 - \frac{1}{4 \cdot 12}$

Step 2.4.2.1.2

Multiply $4$ by $12$ .

$e = 0 - \frac{1}{48}$

Step 2.4.2.2

Subtract $\frac{1}{48}$ from $0$ .

$e = - \frac{1}{48}$

Step 2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $12 {(x + \frac{1}{24})}^{2} - \frac{1}{48}$ .

$12 {(x + \frac{1}{24})}^{2} - \frac{1}{48}$

Step 3

Substitute $12 {(x + \frac{1}{24})}^{2} - \frac{1}{48}$ for $12 x^{2} + x$ in the equation $12 x^{2} - 12 y^{2} + x - y = 3$ .

$12 {(x + \frac{1}{24})}^{2} - \frac{1}{48} - 12 y^{2} - y = 3$

Step 4

Move $- \frac{1}{48}$ to the right side of the equation by adding $\frac{1}{48}$ to both sides.

$12 {(x + \frac{1}{24})}^{2} - 12 y^{2} - y = 3 + \frac{1}{48}$

Step 5

Complete the square for $- 12 y^{2} - y$ .

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Step 5.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 12$

$b = - 1$

$c = 0$

Step 5.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 5.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 5.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 1}{2 \cdot - 12}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Multiply $2$ by $- 12$ .

$d = \frac{- 1}{- 24}$

Step 5.3.2.2

Dividing two negative values results in a positive value.

$d = \frac{1}{24}$

Step 5.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 5.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 1)}^{2}}{4 \cdot - 12}$

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Raise $- 1$ to the power of $2$ .

$e = 0 - \frac{1}{4 \cdot - 12}$

Step 5.4.2.1.2

Multiply $4$ by $- 12$ .

$e = 0 - \frac{1}{- 48}$

Step 5.4.2.1.3

Move the negative in front of the fraction.

$e = 0 - - \frac{1}{48}$

Step 5.4.2.1.4

Multiply $- - \frac{1}{48}$ .

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Step 5.4.2.1.4.1

Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{1}{48})$

Step 5.4.2.1.4.2

Multiply $\frac{1}{48}$ by $1$ .

$e = 0 + \frac{1}{48}$

Step 5.4.2.2

Add $0$ and $\frac{1}{48}$ .

$e = \frac{1}{48}$

Step 5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 12 {(y + \frac{1}{24})}^{2} + \frac{1}{48}$ .

$- 12 {(y + \frac{1}{24})}^{2} + \frac{1}{48}$

Step 6

Substitute $- 12 {(y + \frac{1}{24})}^{2} + \frac{1}{48}$ for $- 12 y^{2} - y$ in the equation $12 x^{2} - 12 y^{2} + x - y = 3$ .

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} + \frac{1}{48} = 3 + \frac{1}{48}$

Step 7

Move $\frac{1}{48}$ to the right side of the equation by adding $\frac{1}{48}$ to both sides.

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} = 3 + \frac{1}{48} - \frac{1}{48}$

Step 8

Simplify $3 + \frac{1}{48} - \frac{1}{48}$ .

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Step 8.1

Combine the numerators over the common denominator.

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} = 3 + \frac{1 - 1}{48}$

Step 8.2

Simplify the expression.

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Step 8.2.1

Subtract $1$ from $1$ .

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} = 3 + \frac{0}{48}$

Step 8.2.2

Divide $0$ by $48$ .

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} = 3 + 0$

Step 8.2.3

Add $3$ and $0$ .

$12 {(x + \frac{1}{24})}^{2} - 12 {(y + \frac{1}{24})}^{2} = 3$

Step 9

Divide each term by $3$ to make the right side equal to one.

$\frac{12 {(x + \frac{1}{24})}^{2}}{3} - \frac{12 {(y + \frac{1}{24})}^{2}}{3} = \frac{3}{3}$

Step 10

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x + \frac{1}{24})}^{2}}{\frac{1}{4}} - \frac{{(y + \frac{1}{24})}^{2}}{\frac{1}{4}} = 1$