# How do you solve (x+2) / (2x+3) - (x-2) / (2x+3) = 21 / (4x^2-9)?

Jul 4, 2016

Shown in a $\underline{\text{lot of detail}}$. With practice you would be able to answer this question type using shortcuts and skipping steps.

$x = \frac{33}{8}$

#### Explanation:

As the bottom values (denominators) are the same on the left of the equals we can write the equation as:

$\frac{\left(x + 2\right) - \left(x - 2\right)}{2 x + 3} = \frac{21}{4 {x}^{2} - 9}$
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$\textcolor{b r o w n}{\text{Consider the left hand side (LHS)}}$

$\frac{\left(x + 2\right) - \left(x - 2\right)}{2 x + 3}$

Multiply the contents $\left(x - 2\right) \text{ by } \left(- 1\right)$ and remove the brackets

$\frac{x + 2 - x + 2}{2 x + 3} \to \frac{4}{2 x + 3}$

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$\textcolor{b r o w n}{\text{Consider the right hand side (RHS)}}$

Write as: $\frac{21}{{\left(2 x\right)}^{2} - {3}^{2}}$

Compare the denominator to ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
We have this condition so write the RHS as

$\frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)}$
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$\textcolor{b r o w n}{\text{Putting it all together}}$

$\frac{4}{2 x + 3} = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)}$
'..........................................................................
$\textcolor{w h i t e}{.}$

Subtract $\textcolor{b l u e}{\frac{4}{2 x + 3}}$ from both sides

$\textcolor{w h i t e}{.}$

$0 = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)} - \frac{4}{2 x + 3}$

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Multiply $\frac{4}{2 x + 3}$ by 1 but in the form of $\textcolor{b l u e}{1 = \frac{2 x - 3}{2 x - 3}}$ giving:

$\textcolor{b r o w n}{0 = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)} - \left[\frac{4}{2 x + 3} \textcolor{b l u e}{\times \frac{2 x - 3}{2 x - 3}}\right]}$

$0 = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)} - \frac{4 \left(2 x - 3\right)}{\left(2 x + 3\right) \left(2 x - 3\right)}$

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They both have the same denominator so combine the fractions.
Jumping steps now.

$0 = \frac{21 - 4 \left(2 x - 3\right)}{\left(2 x - 3\right) \left(2 x + 3\right)}$

$21 - 8 x + 12 = 0$

$x = \frac{33}{8}$
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Jul 4, 2016

$x = \frac{33}{8}$

#### Explanation:

$\frac{x + 2}{2 x + 3} - \frac{x - 2}{2 x + 3} = \frac{21}{4 {x}^{2} - 9}$

Factorise the denominator on the RHS

$\frac{x + 2}{2 x + 3} - \frac{x - 2}{2 x + 3} = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)}$

The following step is explained very clearly by Tony B .. refer to his answer for details.

The denominators on the left are the same so the two fractions can be written as one:

$\frac{x + 2 - \left(x - 2\right)}{2 x + 3} = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)}$

$\frac{x + 2 - x + 2}{2 x + 3} = \frac{21}{\left(2 x + 3\right) \left(2 x - 3\right)}$

$\text{ "4/((2x+3) )= 21/((2x+3)(2x-3)) " cross multiply}$

$4 \left(2 x + 3\right) \left(2 x - 3\right) = 21 \left(2 x + 3\right) \text{ divide by (2x+3)}$

$\frac{4 \cancel{2 x + 3} \left(2 x - 3\right)}{\cancel{\left(2 x + 3\right)}} = \frac{21 \cancel{2 x + 3}}{\cancel{2 x + 3}}$

$8 x - 12 = 21$

$8 x = 33$

$x = \frac{33}{8}$