5|3+7m|+1=51
1) Let's solve in ℝ: 5·|3 + 7m| + 1 = 51
ⓘ We subtract 1 from both sides
5·|3 + 7m| = 51 -1
ⓘ Start from 51 and move back 1 on the number line: 51 -1 = 50
5·|3 + 7m| = 50
Since 5 ≠ 0, divide both sides by 5: |3 + 7m| = 10
Since 10 ≥ 0, |3 + 7m| = 10 is equivalent to 3 + 7m = 10 or 3 + 7m = -10
2) Let's solve in ℝ: 3 + 7m = 10
ⓘ We subtract 3 from both sides
7m = 10 -3
ⓘ Start from 10 and move back 3 on the number line: 10 -3 = 7
7m = 7
ⓘ Both sides are divided by 7
m = 77
ⓘ We simplify the fraction by dividing the numerator and the denominator by 7, their greatest common divisor: 77 = 1
Solution found: m = 1
3) Let's solve in ℝ: 3 + 7m = -10
ⓘ We subtract 3 from both sides
7m = -10 -3
ⓘ Start from -10 and move back 3 more on the number line, away from zero: -10 -3 = -13
7m = -13
ⓘ Both sides are divided by 7
Solution obtained: m = -137
▶Solution:   m = { -137, 1 }
4) The numeric value of: m = { -137, 1 } equals:
▷ m = { ≈ -1.8571, 1 }