$[13 + (- 12)] + [] = [] + [(- 12) + (- 7)]$

$0$

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$20$

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Solution

For making both LHS = RHS we will solve both sides Together
So, $[13 + (- 12)] + [] = [] + [(- 12) + (- 7)]$
$\Rightarrow [13 - 12] + [] = [] + [- 12 - 7] \Rightarrow 1 + [] = [] + [- 19] \Rightarrow 1 + [] = [] - 19$
Here $0$ must be added to LHS and 19 must be subtracted from $20$ so as to get result same both side.

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Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

Simplify: $[12 (104 - 2) + 36 \times 2] \div [13 (10 + 2) - 12]$

A = 13 [12 + {32 \times 5 + (32 - 12 \div 3)}]

B = 13 {- 15 + (13) \times 4}

A \div B

x

{(\frac{12}{13})}^{3 x} \div {(\frac{13}{12})}^{4} = {(\frac{12}{13})}^{10}

(- 13) x [(- 6) + (- 19)] = (- 13) x (- 6) + (- 13) x (- 19)

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