Verify each of the following:

Question:

Verify each of the following:

(i) $\frac{3}{7} \times \frac{-5}{9}=\frac{-5}{9} \times \frac{3}{7}$

(ii) $\frac{-8}{7} \times \frac{13}{9}=\frac{13}{9} \times \frac{-8}{7}$

(iii) $\frac{-12}{5} \times \frac{7}{-36}=\frac{7}{-36} \times \frac{-12}{5}$

(iv) $-8 \times \frac{-13}{12}=\frac{-13}{12} \times(-8)$

 

Solution:

(i) $\frac{3}{7} \times \frac{-5}{9}=\frac{-5}{9} \times \frac{3}{7}$

$\mathrm{LHS}=\frac{3 \times(-5)}{7 \times 9}$

$=-\frac{15}{63}$

Simplifying, we get:

$-\frac{15}{63}=-\frac{15 \div 3}{63 \div 3}$

$=-\frac{5}{21}$

$\mathrm{RHS}=\frac{-5}{9} \times \frac{3}{7}$

$=\frac{(-5) \times 3}{9 \times 7}$

$=\frac{-15}{63}$

Simplifying, we get:

$=\frac{-15 \div 3}{63 \div 3}$

$=-\frac{5}{21}$

LHS = RHS

(ii) $\frac{-8}{7} \times \frac{13}{9}=\frac{13}{9} \times \frac{-8}{7}$

$\mathrm{LHS}=\frac{-8}{7} \times \frac{13}{9}=\frac{(-8) \times 13}{7 \times 9}=-\frac{104}{63} \mathrm{RHS}=\frac{13}{9} \times \frac{-8}{7}=\frac{13 \times(-8)}{9 \times 7}=-\frac{104}{63} \mathrm{LHS}=\mathrm{RHS}$

(iii) $\frac{-12}{5} \times \frac{7}{-36}=\frac{7}{-36} \times \frac{-12}{5}$

$\mathrm{LHS}=\frac{-12}{5} \times \frac{7}{-36}$

$=\frac{(-12) \times 7}{5 \times(-36)}$

$=\frac{84}{180}$

Simplifying, we get:

$=\frac{84 \div 12}{180 \div 12}$

$=\frac{7}{15}$

RHS $=\frac{7}{-36} \times \frac{-12}{5}$

$=\frac{7 \times(-12)}{(-36) \times 5}$

$=\frac{84}{180}$

Simplifying, we get:

$=\frac{84 \div 12}{180 \div 12}$

$=\frac{7}{15}$

LHS = RHS

(iv) $-8 \times \frac{-13}{12}=\frac{-13}{12} \times(-8)$

$\mathrm{LHS}=-8 \times \frac{-13}{12}$

$=\frac{(-8) \times(-13)}{12}$

$=\frac{104}{12}$

Simplifying, we get:

$=\frac{104 \div 4}{12 \div 4}$

$=\frac{26}{3}$

RHS $=\frac{-13}{12} \times(-8)$

$=\frac{(-13) \times(-8)}{12}$

$=\frac{104}{12}$

Simplifying, we get:

$=\frac{104 \div 4}{12 \div 4}$

$=\frac{26}{3}$

LHS = RHS

 

 

 

 

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