Question

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

Answer 1

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]

Answer 2

`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"`