How do you subtract $\frac{c + 2}{3} - \frac{c - 4}{4}$ $\frac { c + 2} { 3} - \frac { c - 4} { 4}$ ?

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Answer 1 · 2018-04-24T10:28:47

Take the constant terms together and the terms having ' $c$ $c$ ' together :-

$\frac{c + 2}{3} - \frac{c - 4}{4} = {\frac{c}{3}} - {\frac{c}{4}} + \frac{2}{3} + \frac{4}{4}$ ${c+2}/3-(c-4)/4={c/3}-{c/4}+2/3+4/4$

$= \frac{c}{12} + \frac{5}{3} = \frac{c + 20}{12}$ $=c/12+5/3=(c+20)/12$

How do you subtract \frac { c + 2} { 3} - \frac { c - 4} { 4}c+23−c−44\frac { c + 2} { 3} - \frac { c - 4} { 4}?