How do you solve # 6/(x+4 )+ 3/4 = (2x+1) / (3x+12)#?

Your starting expression looks like this

#6/(x+4) + 3/4 = (2x+1)/(3x+12)#

Notice that you can use #3# as a common factor for the denominator of the fraction that's on the right-hand side of the equation

#3x + 12 = 3 * (x + 4)#

The equation can thus be written as

#6/(x+4) + 3/4 = (2x+1)/(3(x+4))#

Next, you need to get rid of the denominators. To do that, you need to find the least common multiple of the expressions that act as denominators for your three fractions

#x+4" "#, #" "4" "#, and #" "3(x+4)#

Notice that you need to multiply the first one by #3# and by #4#, the second one by #3# and #(x+4)#, and the third one by #4# to get

#3 * 4 * (x+4)" "#

This will be your least common multiple. The equation will become

#6/(x+4) * 12/12 + 3/4 * (3(x+4))/(3(x+4)) = (2x+1)/(3(x+4)) * 4/4#

#(6 * 12)/(12(x+4)) + (3 * 3(x+4))/(12(x+4)) = ((2x+1) * 4)/(12(x+4))#

This is equivalent to

#72 + 9 * (x+4) = 4 * (2x + 1)#

Finally, expand the parantheses and isolate #x# on one side of the equation

#72 + 9x + 36 = 8x + 4#

#9x - 8x = 4 - 108#

#x = color(green)(-104)#

Do a quick check to make sure the calculations are correct

#6/(-104 + 4) + 3/4 = (2 * (-104) + 1)/(3 * (-104 + 4)#

#-0.06 + 0.75 = 0.69color(white)(x)color(green)(sqrt())#