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

2 Answers
Jul 4, 2016

Answer:

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)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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)#

,.....................................................................................
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#
,.............................................................................................

Jul 4, 2016

Answer:

#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#