Finite Math Examples

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

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

Step 1

Rewrite the equation as

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

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

Step 2

Factor each term.

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Step 2.1

Rewrite

$1$ as

$1^{2}$ .

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

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = x$ and

$b = 1$ .

$\frac{(x + 1) (x - 1)}{x^{2} - 7 x + 12} = y$

Step 2.3

Factor

$x^{2} - 7 x + 12$ using the AC method.

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Step 2.3.1

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

$12$ and whose sum is

$- 7$ .

$- 4, - 3$

Step 2.3.2

Write the factored form using these integers.

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

Step 3

Find the LCD of the terms in the equation.

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Step 3.1

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

$(x - 4) (x - 3), 1$

Step 3.2

The LCM of one and any expression is the expression.

$(x - 4) (x - 3)$

Step 4

Multiply each term in

$\frac{(x + 1) (x - 1)}{(x - 4) (x - 3)} = y$ by

$(x - 4) (x - 3)$ to eliminate the fractions.

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Step 4.1

Multiply each term in

$\frac{(x + 1) (x - 1)}{(x - 4) (x - 3)} = y$ by

$(x - 4) (x - 3)$ .

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of

$(x - 4) (x - 3)$ .

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Step 4.2.1.1

Cancel the common factor.

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

Step 4.2.1.2

Rewrite the expression.

$(x + 1) (x - 1) = y ((x - 4) (x - 3))$

Step 4.2.2

Expand

$(x + 1) (x - 1)$ using the FOIL Method.

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Step 4.2.2.1

Apply the distributive property.

$x (x - 1) + 1 (x - 1) = y ((x - 4) (x - 3))$

Step 4.2.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 (x - 1) = y ((x - 4) (x - 3))$

Step 4.2.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 = y ((x - 4) (x - 3))$

Step 4.2.3

Simplify and combine like terms.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

Multiply

$x$ by

$x$ .

$x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.1.2

Move

$- 1$ to the left of

$x$ .

$x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.1.3

Rewrite

$- 1 x$ as

$- x$ .

$x^{2} - x + 1 x + 1 \cdot - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.1.4

Multiply

$x$ by

$1$ .

$x^{2} - x + x + 1 \cdot - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.1.5

Multiply

$- 1$ by

$1$ .

$x^{2} - x + x - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.2

Add

$- x$ and

$x$ .

$x^{2} + 0 - 1 = y ((x - 4) (x - 3))$

Step 4.2.3.3

Add

$x^{2}$ and

$0$ .

$x^{2} - 1 = y ((x - 4) (x - 3))$

Step 4.3

Simplify the right side.

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Step 4.3.1

Expand

$(x - 4) (x - 3)$ using the FOIL Method.

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Step 4.3.1.1

Apply the distributive property.

$x^{2} - 1 = y (x (x - 3) - 4 (x - 3))$

Step 4.3.1.2

Apply the distributive property.

$x^{2} - 1 = y (x \cdot x + x \cdot - 3 - 4 (x - 3))$

Step 4.3.1.3

Apply the distributive property.

$x^{2} - 1 = y (x \cdot x + x \cdot - 3 - 4 x - 4 \cdot - 3)$

Step 4.3.2

Simplify and combine like terms.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Multiply

$x$ by

$x$ .

$x^{2} - 1 = y (x^{2} + x \cdot - 3 - 4 x - 4 \cdot - 3)$

Step 4.3.2.1.2

Move

$- 3$ to the left of

$x$ .

$x^{2} - 1 = y (x^{2} - 3 \cdot x - 4 x - 4 \cdot - 3)$

Step 4.3.2.1.3

Multiply

$- 4$ by

$- 3$ .

$x^{2} - 1 = y (x^{2} - 3 x - 4 x + 12)$

Step 4.3.2.2

Subtract

$4 x$ from

$- 3 x$ .

$x^{2} - 1 = y (x^{2} - 7 x + 12)$

Step 4.3.3

Apply the distributive property.

$x^{2} - 1 = y x^{2} + y (- 7 x) + y \cdot 12$

Step 4.3.4

Simplify.

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Step 4.3.4.1

Rewrite using the commutative property of multiplication.

$x^{2} - 1 = y x^{2} - 7 y x + y \cdot 12$

Step 4.3.4.2

Move

$12$ to the left of

$y$ .

$x^{2} - 1 = y x^{2} - 7 y x + 12 \cdot y$

$x^{2} - 1 = y x^{2} - 7 y x + 12 y$

Step 5

Solve the equation.

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Step 5.1

Since

$x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$y x^{2} - 7 y x + 12 y = x^{2} - 1$

Step 5.2

Subtract

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

$y x^{2} - 7 y x + 12 y - x^{2} = - 1$

Step 5.3

Add

$1$ to both sides of the equation.

$y x^{2} - 7 y x + 12 y - x^{2} + 1 = 0$

Step 5.4

Use the quadratic formula to find the solutions.

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

Step 5.5

Substitute the values

$a = y - 1$ ,

$b = - 7 y$ , and

$c = 12 y + 1$ into the quadratic formula and solve for

$x$ .

$\frac{7 y \pm \sqrt{{(- 7 y)}^{2} - 4 \cdot ((y - 1) \cdot (12 y + 1))}}{2 (y - 1)}$

Step 5.6

Simplify the numerator.

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Step 5.6.1

Apply the product rule to

$- 7 y$ .

$x = \frac{7 y \pm \sqrt{{(- 7)}^{2} y^{2} - 4 \cdot (y - 1) \cdot (12 y + 1)}}{2 (y - 1)}$

Step 5.6.2

Raise

$- 7$ to the power of

$2$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 \cdot (y - 1) \cdot (12 y + 1)}}{2 (y - 1)}$

Step 5.6.3

Apply the distributive property.

$x = \frac{7 y \pm \sqrt{49 y^{2} + (- 4 y - 4 \cdot - 1) \cdot (12 y + 1)}}{2 (y - 1)}$

Step 5.6.4

Multiply

$- 4$ by

$- 1$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} + (- 4 y + 4) \cdot (12 y + 1)}}{2 (y - 1)}$

Step 5.6.5

Expand

$(- 4 y + 4) (12 y + 1)$ using the FOIL Method.

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Step 5.6.5.1

Apply the distributive property.

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 y (12 y + 1) + 4 (12 y + 1)}}{2 (y - 1)}$

Step 5.6.5.2

Apply the distributive property.

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 y (12 y) - 4 y \cdot 1 + 4 (12 y + 1)}}{2 (y - 1)}$

Step 5.6.5.3

Apply the distributive property.

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 y (12 y) - 4 y \cdot 1 + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6

Simplify and combine like terms.

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Step 5.6.6.1

Simplify each term.

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Step 5.6.6.1.1

Rewrite using the commutative property of multiplication.

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 \cdot (12 y \cdot y) - 4 y \cdot 1 + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.2

Multiply

$y$ by

$y$ by adding the exponents.

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Step 5.6.6.1.2.1

Move

$y$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 \cdot (12 (y \cdot y)) - 4 y \cdot 1 + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.2.2

Multiply

$y$ by

$y$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 4 \cdot (12 y^{2}) - 4 y \cdot 1 + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.3

Multiply

$- 4$ by

$12$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 48 y^{2} - 4 y \cdot 1 + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.4

Multiply

$- 4$ by

$1$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 48 y^{2} - 4 y + 4 (12 y) + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.5

Multiply

$12$ by

$4$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 48 y^{2} - 4 y + 48 y + 4 \cdot 1}}{2 (y - 1)}$

Step 5.6.6.1.6

Multiply

$4$ by

$1$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 48 y^{2} - 4 y + 48 y + 4}}{2 (y - 1)}$

Step 5.6.6.2

Add

$- 4 y$ and

$48 y$ .

$x = \frac{7 y \pm \sqrt{49 y^{2} - 48 y^{2} + 44 y + 4}}{2 (y - 1)}$

Step 5.6.7

Subtract

$48 y^{2}$ from

$49 y^{2}$ .

$x = \frac{7 y \pm \sqrt{y^{2} + 44 y + 4}}{2 (y - 1)}$

Step 5.7

The final answer is the combination of both solutions.

$x = \frac{7 y + \sqrt{y^{2} + 44 y + 4}}{2 (y - 1)}$

$x = \frac{7 y - \sqrt{y^{2} + 44 y + 4}}{2 (y - 1)}$

$x = \frac{7 y + \sqrt{y^{2} + 44 y + 4}}{2 (y - 1)}$

$x = \frac{7 y - \sqrt{y^{2} + 44 y + 4}}{2 (y - 1)}$