Subtract: (i) -18/11 from 1 (ii) -13/9 from 0 (iii) -32/13 from -6/5 (iv) -7 from -4/7

Question

Subtract:

 (i) \(\frac{-18}{11}\) from 1

(ii) \(\frac{-13}{9}\) from 0

(iii) \(\frac{-32}{13}\) from \(\frac{-6}{5}\) 

(iv) \(-7\) from \(\frac{-4}{7}\) 

Answer 1

(i) 

 We can write, 1 = \(\frac{1}{1}\)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 1 and 11 = 11

\(\frac{1}{1} = \frac{1\times11}{1\times11}= \frac{11}{11}\)

And,

\(\frac{18}{11} = \frac{-18\times1}{11\times1}= \frac{-18}{11}\)

Therefore,

\(1-(\frac{-18}{11})\)

= \(\frac{11}{11}-(\frac{-18}{11})\) 

= \(\frac{11-(-18)}{11}\) 

= \(\frac{11+18}{11}\)

= \(\frac{29}{11}\)

(ii) 

\(0-(\frac{-13}{9})\)

= \(0+\frac{13}{9}\)

= \(\frac{13}{9}\)

(iii)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 13 and 5 = 65

\(\frac{-6}{5}= \frac{-6\times13}{5\times13}\frac{-78}{65}\)

Therefore,

\(\frac{-6}{5}-(\frac{-32}{13})\)

= \(\frac{-78}{65}-(\frac{-160}{65})\)

= \(\frac{-78-(-160)}{65}\) 

= \(\frac{-78+160}{65}\)

= \(\frac{82}{65}\)

(iv)

We can write, \(-7= \frac{-7}{1}\)

Since the denominators of both the numbers are different 

therefore, 

we will take their LCM 0f 1 and 7 = 7

\(\frac{-7}{1}= \frac{-7\times7}{1\times7}= \frac{-49}{7}\)

And,

\(\frac{-4}{7}= \frac{-4\times1}{7\times1}= \frac{-4}{7}\)

Therefore,

= \(\frac{-4}{7}-(-7)\) 

= \(\frac{-4}{7}-(\frac{-49}{7})\)

= \(\frac{-4-(-49)}{7}\) 

= \(\frac{-4+49}{7}\) 

= \(\frac{45}{7}\)