How do you evaluate #\frac { 2x + 1} { 2x - 3} - \frac { 7x } { 4x ^ { 2} - 9} = 1+ \frac { x - 4} { 2x + 3}#?

1 Answer
Noah G
Jan 2, 2017

#x= 0 and 6#

Explanation:

The first step to solving rational equations is determining the LCD (Least Common Denominator).

#4x^2 - 9 = (2x + 3)(2x - 3)#

#:.#The LCD is #(2x + 3)(2x - 3)#

Once you know this, you have to put each fraction on this denominator.

#((2x + 1)(2x + 3))/((2x- 3)(2x + 3)) - (7x)/((2x + 3)(2x- 3)) = (1(2x- 3)(2x + 3))/((2x + 3)(2x -3)) + ((x- 4)(2x - 3))/((2x - 3)(2x + 3))#

We can now eliminate denominators, since all fractions are now equivalent.

#(2x + 1)(2x + 3) - 7x = (2x - 3)(2x + 3) + (x - 4)(2x - 3)#

Expand:

#4x^2 + 2x + 6x + 3 - 7x = 4x^2 - 6x + 6x - 9 + 2x^2 - 8x - 3x + 12#

Combine like terms.

#0 = 2x^2 - x - 11x#

#0 = 2x^2 - 12x#

#0 = 2x(x- 6)#

#x = 0 and 6#

Always make sure neither solution contradict the restrictions on the variable. If they do,they are extraneous. In this case, restrictions are #x != +- -3/2#. So, our solutions are good. Restrictions will occur whenever the denominator equals #0#, or becomes undefined in the real number system.

Hopefully this helps!

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