How do you solve #x /( 2x + 1 ) + ( 1 / 4 ) = 2 / (2x + 1)# and find any extraneous solutions?

2 Answers
Nov 7, 2017

Answer:

Solution: #1 1/6# and no extraneous root.

Explanation:

#x/(2x+1)+1/4= 2/(2x+1)# Multiplying by #4(2x+1)# on

both sides we get , #4x+(2x+1)= 8 or 6x= 7or x=7/6#

Check: L.H.S #=x/(2x+1)+1/4= (7/6)/(2*7/6+1) +1/4#

#= (7/cancel6)/(20/cancel6)+1/4#

#=7/20+1/4 = 12/20= 3/5#

R.H.S #=2/(2x+1) = 2/(2*7/6+1)=2/(20/6)=12/20=3/5#

L.H.S=R.H.S , no extraneous root.

Solution: #x= 7/6= 1 1/6# [Ans]

Nov 7, 2017

Answer:

#x=7/6#

Explanation:

#"since fractions on both sides of the equation have a"#
#color(blue)"common denominator "" we can combine them"#

#"subtract "2/(2x+1)" from both sides"#

#x/(2x+1)-2/(2x+1)+1/4=0#

#"subtract "1/4" from both sides"#

#x/(2x+1)-2/(2x+1)cancel(+1/4)cancel(-1/4)=0-1/4#

#rArr(x-2)/(2x+1)=-1/4larrcolor(blue)"combining fractions"#

#"using the method of "color(blue)"cross-multiplying"#

#•color(white)(x)a/b=c/drArraxxd=bxxc#

#"attaching the negative sign to 1 "#

#rArr-1(2x+1)=4(x-2)#

#rArr-2x-1=4x-8larrcolor(blue)"distributing"#

#"subtract 4x from both sides"#

#-2x-4x-1=cancel(4x)cancel(-4x)-8#

#rArr-6x-1=-8#

#"add 1 to both sides"#

#-6xcancel(-1)cancel(+1)=-8+1#

#rArr-6x=-7#

#"divide both sides by "-6#

#(cancel(-6) x)/cancel(-6)=(-7)/(-6)#

#rArrx=7/6" is the solution"#

#"there are no extraneous solutions"#