Precalculus Examples

$\frac{7 x - 2}{x^{2} + 8} = 1$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

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

$x^{2} + 8, 1$

Step 1.2

Remove parentheses.

$x^{2} + 8, 1$

Step 1.3

The LCM of one and any expression is the expression.

$x^{2} + 8$

Step 2

Multiply each term in $\frac{7 x - 2}{x^{2} + 8} = 1$ by $x^{2} + 8$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{7 x - 2}{x^{2} + 8} = 1$ by $x^{2} + 8$ .

$\frac{7 x - 2}{x^{2} + 8} (x^{2} + 8) = 1 (x^{2} + 8)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $x^{2} + 8$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{7 x - 2}{x^{2} + 8} (x^{2} + 8) = 1 (x^{2} + 8)$

Step 2.2.1.2

Rewrite the expression.

$7 x - 2 = 1 (x^{2} + 8)$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply $x^{2} + 8$ by $1$ .

$7 x - 2 = x^{2} + 8$

Step 3

Solve the equation.

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Step 3.1

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

$7 x - 2 - x^{2} = 8$

Step 3.2

Subtract $8$ from both sides of the equation.

$7 x - 2 - x^{2} - 8 = 0$

Step 3.3

Subtract $8$ from $- 2$ .

$7 x - x^{2} - 10 = 0$

Step 3.4

Factor the left side of the equation.

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Step 3.4.1

Factor $- 1$ out of $7 x - x^{2} - 10$ .

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Step 3.4.1.1

Reorder $7 x$ and $- x^{2}$ .

$- x^{2} + 7 x - 10 = 0$

Step 3.4.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 7 x - 10 = 0$

Step 3.4.1.3

Factor $- 1$ out of $7 x$ .

$- (x^{2}) - (- 7 x) - 10 = 0$

Step 3.4.1.4

Rewrite $- 10$ as $- 1 (10)$ .

$- (x^{2}) - (- 7 x) - 1 \cdot 10 = 0$

Step 3.4.1.5

Factor $- 1$ out of $- (x^{2}) - (- 7 x)$ .

$- (x^{2} - 7 x) - 1 \cdot 10 = 0$

Step 3.4.1.6

Factor $- 1$ out of $- (x^{2} - 7 x) - 1 (10)$ .

$- (x^{2} - 7 x + 10) = 0$

Step 3.4.2

Factor.

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Step 3.4.2.1

Factor $x^{2} - 7 x + 10$ using the AC method.

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Step 3.4.2.1.1

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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Step 3.4.2.1.2

Write the factored form using these integers.

$- ((x - 5) (x - 2)) = 0$

Step 3.4.2.2

Remove unnecessary parentheses.

$- (x - 5) (x - 2) = 0$

Step 3.5

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

$x - 5 = 0$

$x - 2 = 0$

Step 3.6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 3.6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 3.7

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 3.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 3.8

The final solution is all the values that make $- (x - 5) (x - 2) = 0$ true.

$x = 5, 2$