(2m+1)/4+3= m/6-(6-m)/(12) — step-by-step solution

Let's solve in

ℝ

\frac{2 m + 1}{4} + 3 = \frac{m}{6} - \frac{6 - m}{12}

ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- \frac{6 - m}{12}

\frac{- 6 + m}{12}

ⅰ

\frac{2 m + 1}{4} + 3

\frac{(2 m + 1) + 4 \cdot 3}{4}

\frac{(2 m + 1) + 4 \cdot 3}{4} = \frac{m}{6} + \frac{- 6 + m}{12}

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 3

12

■To add fractions, put them over the same denominator, then add the numerators:

\frac{m}{6} + \frac{- 6 + m}{12}

\frac{2 m + (- 6 + m)}{12}

\frac{2 m + 1 + 12}{4} = \frac{2 m + (- 6 + m)}{12}

ⓘ The numerator and the denominator are computed before the division:

1 + 12

13

■Like terms are grouped by adding their coefficients:

2 m + m

3 m

\frac{2 m + 13}{4} = \frac{3 m - 6}{12}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{3 m - 6}{12}

\frac{m - 2}{4}

\frac{2 m + 13}{4} = \frac{m - 2}{4}

Let's sum the fractions:

\frac{2 m + 13}{4} - \frac{m - 2}{4} = 0

ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- \frac{m - 2}{4}

\frac{- m + 2}{4}

■To add fractions, put them over the same denominator, then add the numerators:

\frac{2 m + 13}{4} + \frac{- m + 2}{4}

\frac{(2 m + 13) + (- m + 2)}{4}

\frac{(2 m + 13) + (- m + 2)}{4} = 0

ⓘ The numerator and the denominator are computed before the division:

13 + 2

15

■Like terms are grouped by adding their coefficients:

2 m - m

m

\frac{m + 15}{4} = 0

⇥Let's solve in

ℝ

m + 15 = 0

▶Solution obtained:

m = -15