We want to solve the given rational equation.

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

Let's begin by highlighting the all of the different factors in the denominators. This will help us find the least common denominator (LCD).

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x} ⇕ \frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

We will solve the equation by multiplying each side by the LCD of the denominators.

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

MultEqn

$LHS \cdot 2 x = RHS \cdot 2 x$

(\frac{1}{x} + \frac{x}{2}) \cdot 2 x = (\frac{x + 4}{2 x}) \cdot 2 x

Multiply

\frac{2 x}{x} + \frac{x ( 2 x )}{2} = \frac{( x + 4 ) ( 2 x )}{2 x}

CancelCommonFac

Cancel out common factors

\frac{2 x}{x} + \frac{x ( 2 x )}{2} = \frac{( x + 4 ) 2 x}{2 x}

SimpQuot

Simplify quotient

2 + x (x) = x + 4

Multiply

2 + x^{2} = x + 4

▼

Simplify

SubEqn

$LHS - x = RHS - x$

2 + x^{2} - x = 4

SubEqn

$LHS - 4 = RHS - 4$

x^{2} - x - 2 = 0

Note that we have a quadratic equation now. Let's identify the values of

a,

b,

and

c .

x^{2} - x - 2 = 0 \Leftrightarrow 1 x^{2} + (- 1) x + (- 2) = 0

We have that

a = 1,

b = - 1,

and

c = - 2 .

Let's substitute these values into the Quadratic Formula and solve for

x .

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 2 )}{2 ( 1 )}

▼

Simplify right-hand side

NegNeg

$- (- a) = a$

x = \frac{1 \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 2 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

x = \frac{1 \pm 1 + 8}{2}

AddTerms

Add terms

x = \frac{1 \pm 9}{2}

CalcRoot

Calculate root

x = \frac{1 \pm 3}{2}

Let's calculate both solutions by using the positive and negative signs.

$x = \frac{1 \pm 3}{2}$
$x = \frac{1 + 3}{2}$	$x = \frac{1 - 3}{2}$
$x = \frac{4}{2}$	$x = \frac{- 2}{2}$
$x = 2$	$x = - 1$

Finally, we will check if either

x = 2

or

x = - 1

is an extraneous solution. To do this, we need to substitute

2

and

- 1

for

x

in the original equation. Let's start with

x = 2 .

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

Substitute

$x = 2$

\frac{1}{2} + \frac{2}{2} = ? \frac{2 + 4}{2 ( 2 )}

▼

Simplify

AddTerms

Add terms

\frac{1}{2} + \frac{2}{2} = ? \frac{6}{2 ( 2 )}

Multiply

\frac{1}{2} + \frac{2}{2} = ? \frac{6}{4}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{2} + \frac{2}{2} = ? \frac{3}{2}

AddFrac

Add fractions

\frac{3}{2} = \frac{3}{2} ✓

Now, we will check our second solution,

x = - 1 .

\frac{1}{x} + \frac{x}{2} = \frac{x + 4}{2 x}

Substitute

$x = - 1$

\frac{1}{- 1} + \frac{- 1}{2} = ? \frac{- 1 + 4}{2 ( - 1 )}

▼

Simplify

AddTerms

Add terms

\frac{1}{- 1} + \frac{- 1}{2} = ? \frac{3}{2 ( - 1 )}

MultPosNeg

$a (- b) = - a \cdot b$

\frac{1}{- 1} + \frac{- 1}{2} = ? \frac{3}{- 2}

MoveNegDenomToFrac

Put minus sign in front of fraction

- \frac{1}{1} - \frac{1}{2} = ? - \frac{3}{2}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

- \frac{2}{2} - \frac{1}{2} = ? - \frac{3}{2}

SubFrac

Subtract fractions

- \frac{3}{2} = - \frac{3}{2} ✓

Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are

x = 2

and

x = - 1 .

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