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

Question

Savannah Anderson · Accepted Answer

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

#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# No superfluous root; L.H.S.=R.H.S. Solution: #x= 7/6= 1 1/6# [Ans]

Abigail Allen · Answer

#x=7/6# #"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"#

Julia Adams · Answer

undefined To solve the equation x / (2x + 1) + (1 / 4) = 2 / (2x + 1) and find any extraneous solutions, follow these steps:

1. Start by multiplying both sides of the equation by the common denominator, which is (2x + 1) * 4. This will eliminate the fractions.

2. After multiplying, simplify the equation and combine like terms.

3. Solve the resulting equation for x.

4. Check the solutions obtained by substituting them back into the original equation to identify any extraneous solutions.