1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$

Method I

Change the mixed fractions to improper fractions.

1$\frac{3}{4}$ = $\frac{\mathrm{1\; x\; 4\; +\; 3}}{4}$ = $\frac{\mathrm{4\; +\; 3}}{4}$ = $\frac{7}{4}$

2$\frac{5}{6}$ = $\frac{\mathrm{2\; x\; 6\; +\; 5}}{6}$ = $\frac{\mathrm{12\; +\; 5}}{6}$ = $\frac{17}{6}$

1$\frac{9}{10}$ = $\frac{\mathrm{1\; x\; 10\; +\; 9}}{10}$ = $\frac{\mathrm{10\; +\; 9}}{10}$ = $\frac{19}{10}$

4$\frac{7}{8}$ = $\frac{\mathrm{4\; x\; 8\; +\; 7}}{8}$ = $\frac{\mathrm{32\; +\; 7}}{8}$ = $\frac{39}{8}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = $\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$

The L.C.M of the denominators (4,6,10,8) is 120

You can as well express the denominators in prime factors and find the L.C.M from them.

4 = 2 x 2 = 2²
6 = 2 x 3
10 = 2 x 5
8 = 2 x 2 x 2 = 2³

The L.C.M is the product (multiplication) of each of the prime factors in their highest powers.

For 2, 2³
3 is 3
5 is 5

L.C.M = 2³ x 3 x 5 = 8 x 15 = 120

$\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$ = $\frac{\mathrm{7\; x\; 30\; -\; 20\; x\; 17\; -\; 12\; x\; 19\; +\; 15\; x\; 39}}{120}$

$\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$ = $\frac{\mathrm{210\; -\; 340\; -\; 228\; +\; 585}}{120}$

$\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$ = $\frac{\mathrm{795\; -\; 568}}{120}$

$\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$ = $\frac{227}{120}$

$\frac{7}{4}$ - $\frac{17}{6}$ - $\frac{19}{10}$ + $\frac{39}{8}$ = 1$\frac{107}{120}$

Method II

Solve the whole numbers separately and the fractions separately.

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$

1 - 2 - 1 + 4 = 5 - 3
1 - 2 - 1 + 4 = 2

L.C.M = 120

$\frac{3}{4}$ - $\frac{5}{6}$ - $\frac{9}{10}$ + $\frac{7}{8}$ = $\frac{\mathrm{3\; x\; 30\; -\; 5\; x\; 20\; -\; 9\; x\; 12\; +\; 7\; x\; 15}}{120}$

$\frac{3}{4}$ - $\frac{5}{6}$ - $\frac{9}{10}$ + $\frac{7}{8}$ = $\frac{\mathrm{90\; -\; 100\; -\; 108\; +\; 105}}{120}$

$\frac{3}{4}$ - $\frac{5}{6}$ - $\frac{9}{10}$ + $\frac{7}{8}$ = $\frac{\mathrm{195\; -\; 208}}{120}$

$\frac{3}{4}$ - $\frac{5}{6}$ - $\frac{9}{10}$ + $\frac{7}{8}$ = $\frac{-13}{120}$

Add the whole number part to the fraction.

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = 2 + $\frac{-13}{120}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = $\frac{2}{1}$ - $\frac{13}{120}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = $\frac{\mathrm{2\; x\; 120\; -\; 13}}{120}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = $\frac{\mathrm{240\; -\; 13}}{120}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = $\frac{227}{120}$

1$\frac{3}{4}$ - 2$\frac{5}{6}$ - 1$\frac{9}{10}$ + 4$\frac{7}{8}$ = 1$\frac{107}{120}$

Simplify the ones in the bracket first

Apply the rule of division of fraction

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