Question

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

Answer 1

Shown in a `ul("lot of detail")`. With practice you would be able to answer this question type using shortcuts and skipping steps.

`x=33/8`

Explanation:

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

`((x+2)-(x-2))/(2x+3)=21/(4x^2-9)`
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`color(brown)("Consider the left hand side (LHS)")`

`((x+2)-(x-2))/(2x+3)`

Multiply the contents `(x-2)" by "(-1)` and remove the brackets

`(x+2-x+2)/(2x+3) -> 4/(2x+3)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Consider the right hand side (RHS)")`

Write as: `21/[(2x)^2-3^2]`

Compare the denominator to `a^2-b^2=(a + b)(a - b)`
We have this condition so write the RHS as

`21/((2x+3)(2x-3))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Putting it all together")`

`4/(2x+3)=21/((2x+3)(2x-3))`
'..........................................................................
`color(white)(.)`

Subtract `color(blue)(4/(2x+3))` from both sides

`color(white)(.)`

`0=21/((2x+3)(2x-3)) - 4/(2x+3)`

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Multiply `4/(2x+3)` by 1 but in the form of `color(blue)(1=(2x-3)/(2x-3))` giving:

`color(brown)(0=21/((2x+3)(2x-3)) - [4/(2x+3)color(blue)(xx(2x-3)/(2x-3))])`

`0=21/((2x+3)(2x-3)) - (4(2x-3))/((2x+3)(2x-3))`

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

`0=(21-4(2x-3))/((2x-3)(2x+3))`

`21-8x+12=0`

`x=33/8`
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Answer 2

`x = 33/8`

Explanation:

`(x+2)/(2x+3) - (x-2)/(2x+3) = 21/(4x^2-9)`

Factorise the denominator on the RHS

`(x+2)/(2x+3) - (x-2)/(2x+3) = 21/((2x+3)(2x-3))`

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:

`(x+2-(x-2))/(2x+3) = 21/((2x+3)(2x-3))`

`(x+2-x+2)/(2x+3) = 21/((2x+3)(2x-3))`

`" "4/((2x+3) )= 21/((2x+3)(2x-3)) " cross multiply"`

`4(2x+3)(2x-3) = 21(2x+3) " divide by (2x+3)"`

`(4cancel(2x+3)(2x-3))/cancel((2x+3)) = (21cancel(2x+3))/(cancel(2x+3))`

`8x -12 = 21`

`8x = 33`

`x = 33/8`