How do you solve for u in #4/(u+6)=6/(u+6)+2#?

2 Answers
Jan 11, 2018

Answer:

#u=-7#

Explanation:

First, we need to put the requirements, that is #u!=-6# because then the denominator will be #0#, and make the equation undefined.

Then,

#4/(u+6)=6/(u+6)+2#

#-2/(u+6)=2#

#u+6=-1#

#u=-7#

Feb 8, 2018

Answer:

A different approach: #u=-7#

Explanation:

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

#color(green)(4/(u+6)=6/(u+6)+2 color(white)("d")->color(white)("d")4/(u+6)=6/(u+6)+[2color(red)(xx1)])#

#color(green)( color(white)("ddddddddddddddddd")->color(white)("d")4/(u+6)=6/(u+6)+[2color(red)(xx(u+6)/(u+6))])#

#color(green)(color(white)("dDDDDddddddddddd")->color(white)("d")4/(u+6)=6/(u+6)+color(white)("d")(2(u+6))/(u+6)#

Now all the denominators (bottom numbers) are the same we can forget about them.

Or, as a purist would say: multiply all of both sides by #(u+6)#. This cancels out the denominators which is THE SAME THING!

#color(green)(4=6+2(u+6))#

#color(green)(4=6+2u+12)#

#color(green)(4=18+2u)#

Subtract 18 from both sides

#color(green)(2u=-14)#

Divide both sides by 2

#color(green)(u=-7)#

#color(blue)("Foot note: as "u" can only take on one value for this to work ")##color(blue)("and this is not -6 then we do not need to state that "x!=-6)#