Precalculus Examples

$2 x + 1 = \frac{5}{x + 2}$

Step 1

Subtract $1$ from both sides of the equation.

$2 x = \frac{5}{x + 2} - 1$

Step 2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 2.1

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

$1, x + 2, 1$

Step 2.2

Remove parentheses.

$1, x + 2, 1$

Step 2.3

The LCM of one and any expression is the expression.

$x + 2$

Step 3

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

Tap for more steps...

Step 3.1

Multiply each term in $2 x = \frac{5}{x + 2} - 1$ by $x + 2$ .

$2 x (x + 2) = \frac{5}{x + 2} (x + 2) - (x + 2)$

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Apply the distributive property.

$2 x \cdot x + 2 x \cdot 2 = \frac{5}{x + 2} (x + 2) - (x + 2)$

Step 3.2.2

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 3.2.2.1

Move $x$ .

$2 (x \cdot x) + 2 x \cdot 2 = \frac{5}{x + 2} (x + 2) - (x + 2)$

Step 3.2.2.2

Multiply $x$ by $x$ .

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

Step 3.2.3

Multiply $2$ by $2$ .

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

Step 3.3

Simplify the right side.

Tap for more steps...

Step 3.3.1

Simplify each term.

Tap for more steps...

Step 3.3.1.1

Cancel the common factor of $x + 2$ .

Tap for more steps...

Step 3.3.1.1.1

Cancel the common factor.

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

Step 3.3.1.1.2

Rewrite the expression.

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

Step 3.3.1.2

Apply the distributive property.

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

Step 3.3.1.3

Multiply $- 1$ by $2$ .

$2 x^{2} + 4 x = 5 - x - 2$

Step 3.3.2

Subtract $2$ from $5$ .

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

Step 4

Solve the equation.

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

Add $x$ to both sides of the equation.

$2 x^{2} + 4 x + x = 3$

Step 4.1.2

Add $4 x$ and $x$ .

$2 x^{2} + 5 x = 3$

Step 4.2

Subtract $3$ from both sides of the equation.

$2 x^{2} + 5 x - 3 = 0$

Step 4.3

Factor by grouping.

Tap for more steps...

Step 4.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Tap for more steps...

Step 4.3.1.1

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 3 = 0$

Step 4.3.1.2

Rewrite $5$ as $- 1$ plus $6$

$2 x^{2} + (- 1 + 6) x - 3 = 0$

Step 4.3.1.3

Apply the distributive property.

$2 x^{2} - 1 x + 6 x - 3 = 0$

Step 4.3.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 4.3.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 6 x - 3 = 0$

Step 4.3.2.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x - 1) + 3 (2 x - 1) = 0$

Step 4.3.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 3) = 0$

Step 4.4

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

$2 x - 1 = 0$

$x + 3 = 0$

Step 4.5

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.5.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 4.5.2

Solve $2 x - 1 = 0$ for $x$ .

Tap for more steps...

Step 4.5.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 4.5.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

Tap for more steps...

Step 4.5.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.5.2.2.2

Simplify the left side.

Tap for more steps...

Step 4.5.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 4.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 4.6

Set $x + 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 4.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.7

The final solution is all the values that make $(2 x - 1) (x + 3) = 0$ true.

$x = \frac{1}{2}, - 3$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}, - 3$

Decimal Form:

$x = 0.5, - 3$