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())`