Finite Math Examples

$\frac{4 x + 20}{{(x - 3)}^{2}} = 1$

Step 1

Factor $4$ out of $4 x + 20$ .

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Step 1.1

Factor $4$ out of $4 x$ .

$\frac{4 (x) + 20}{{(x - 3)}^{2}} = 1$

Step 1.2

Factor $4$ out of $20$ .

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

Step 1.3

Factor $4$ out of $4 x + 4 \cdot 5$ .

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

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

${(x - 3)}^{2}, 1$

Step 2.2

The LCM of one and any expression is the expression.

${(x - 3)}^{2}$

Step 3

Multiply each term in $\frac{4 (x + 5)}{{(x - 3)}^{2}} = 1$ by ${(x - 3)}^{2}$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{4 (x + 5)}{{(x - 3)}^{2}} = 1$ by ${(x - 3)}^{2}$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of ${(x - 3)}^{2}$ .

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Step 3.2.1.1

Cancel the common factor.

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

Step 3.2.1.2

Rewrite the expression.

$4 (x + 5) = 1 {(x - 3)}^{2}$

Step 3.2.2

Apply the distributive property.

$4 x + 4 \cdot 5 = 1 {(x - 3)}^{2}$

Step 3.2.3

Multiply $4$ by $5$ .

$4 x + 20 = 1 {(x - 3)}^{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply ${(x - 3)}^{2}$ by $1$ .

$4 x + 20 = {(x - 3)}^{2}$

Step 4

Solve the equation.

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Step 4.1

Simplify ${(x - 3)}^{2}$ .

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Step 4.1.1

Rewrite.

$4 x + 20 = 0 + 0 + {(x - 3)}^{2}$

Step 4.1.2

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$4 x + 20 = (x - 3) (x - 3)$

Step 4.1.3

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

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Step 4.1.3.1

Apply the distributive property.

$4 x + 20 = x (x - 3) - 3 (x - 3)$

Step 4.1.3.2

Apply the distributive property.

$4 x + 20 = x \cdot x + x \cdot - 3 - 3 (x - 3)$

Step 4.1.3.3

Apply the distributive property.

$4 x + 20 = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

Step 4.1.4

Simplify and combine like terms.

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Step 4.1.4.1

Simplify each term.

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Step 4.1.4.1.1

Multiply $x$ by $x$ .

$4 x + 20 = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3$

Step 4.1.4.1.2

Move $- 3$ to the left of $x$ .

$4 x + 20 = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

Step 4.1.4.1.3

Multiply $- 3$ by $- 3$ .

$4 x + 20 = x^{2} - 3 x - 3 x + 9$

Step 4.1.4.2

Subtract $3 x$ from $- 3 x$ .

$4 x + 20 = x^{2} - 6 x + 9$

Step 4.2

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

$x^{2} - 6 x + 9 = 4 x + 20$

Step 4.3

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

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Step 4.3.1

Subtract $4 x$ from both sides of the equation.

$x^{2} - 6 x + 9 - 4 x = 20$

Step 4.3.2

Subtract $4 x$ from $- 6 x$ .

$x^{2} - 10 x + 9 = 20$

Step 4.4

Subtract $20$ from both sides of the equation.

$x^{2} - 10 x + 9 - 20 = 0$

Step 4.5

Subtract $20$ from $9$ .

$x^{2} - 10 x - 11 = 0$

Step 4.6

Factor $x^{2} - 10 x - 11$ using the AC method.

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Step 4.6.1

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 $- 11$ and whose sum is $- 10$ .

$- 11, 1$

Step 4.6.2

Write the factored form using these integers.

$(x - 11) (x + 1) = 0$

Step 4.7

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

$x - 11 = 0$

$x + 1 = 0$

Step 4.8

Set $x - 11$ equal to $0$ and solve for $x$ .

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Step 4.8.1

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Step 4.8.2

Add $11$ to both sides of the equation.

$x = 11$

Step 4.9

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

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Step 4.9.1

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

$x + 1 = 0$

Step 4.9.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.10

The final solution is all the values that make $(x - 11) (x + 1) = 0$ true.

$x = 11, - 1$