How do you solve #5abs(3+7m)+1=51#?

The first thing to do here is isolate the modules on one side of the equation. To do that, subtract #1# from both sides of the equation

#5 * |3 + 7m| + color(red)(cancel(color(black)(1))) - 1 = 51-1#

#5 * |3 + 7m| = 50#

Now divide both sides of the equation by #5# to get

#(color(red)(cancel(color(black)(5))) * |3 + 7m|)/color(red)(cancel(color(black)(5))) = 50/5#

#|3 + 7m| = 10#

At this point, you have two possibilities to cover

#color(white)(a)#

• #3+7m >=0 implies |3 + 7m| = 3 + 7m#

In this case, the above equation will get you

#3 + 7m = 10#

#7m = 7 implies m = 7/7 = 1#

#color(white)(a)#

• #3 + 7m < 0 implies |3 + 7m| = - (3 + 7m)#

In this case, the equation becomes

#-(3 + 7m) = 10#

#-3 - 7m = 10#

#-7m = 13 implies m = -13/7#

You can thus say that the original equation has two possibile solutions

#m = -13/7" "# or #" "m = 1#

Do a quick check to make sure that the calculations are correct

#5 * |3 + 7 * 1| +1 = 51#

#5 * 10 + 1 = 51 " "color(green)(sqrt())#

#5 * |3 + 7 * (-13/7)| + 1 = 51#

#5 * |3 - 13| + 1 = 51#

#5 * 10 + 1 = 51 " " color(green)(sqrt())#