Question

Question:-
1) By taking x=3/5 and y=4/9, find out whether
$(x + y {)}^{ - 1} = {x}^{ - 1} + {y}^{ - 1}$
is true or false.

2).ADD THESE RATIONAL NUMBERS:-

1)(-2)/5 and 3/4

2)5/8 and (-11)/12

3)7/(-18)+5/(-12)+(-9)/(-16)

Please solve fast in detail please!!!

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Answer 1

GIVEN:-

By taking x=3/5 and y=4/9, find out whether

$(x + y {)}^{ - 1} = {x}^{ - 1} + {y}^{ - 1}$

is true or false.

TO FIND:-

$(x + y {)}^{ - 1} = {x}^{ - 1} + {y}^{ - 1}$ is true or false.

SOLUTION:-

$x + y = \frac{3}{5} + \frac{4}{9} = \frac{3 \times 9 + 4 \times 5}{45} \\ \\ \\ = > \frac{27 + 20}{45} \\ \\ \\ = > \frac{47}{45} \\ \\ \\ (x + y {)}^{ - 1} = \binom{4 {7}^{ - 1} }{45} = \frac{45}{47}. \\ \\ \\ = > {x}^{ - 1} = \binom{ {3}^{ - 1} }{ {5}^{} } \\ \\ \\ = > {y}^{ - 1} = \binom{ {4}^{ - 1} }{9} \\ \\ \\ = > \frac{9}{4} \\ \\ \\ so.. {x}^{ - 1} + {y}^{ - 1} = \frac{5}{3} + \frac{9}{4} \\ \\ \\ = > \frac{5 \times 4 + 9 \times 3}{12} \\ \\ \\ = > \frac{20 + 27}{12} \\ \\ \\ = > \frac{47}{12}. \\ \\ \\ this \: shows \: that \: (x + y {)}^{ - 1}not \: equal \: to \: {x}^{ - 1} + {y}^{ - 1}$

$\small\bold\red{Hence, \: The \: given \: statement \: is \: false!}$

GIVEN:-

Add these rational numbers.

TO FIND:-

Sum of (-2)/5 and 3/4.

SOLUTION:-

The L.C.M. of 5 and 4 is 20.We express the given numbers with common denominator 20.

(-2)/5=(-2)×4/5×4=(-8)/20

3/4=3×5/4×5=15/20.

:. (-2)/5+3/4=(-8)/20+15/20

=>7/20

NOTE THAT: we have obtained the above working (-2)/5+3/4 as (-8)+15/20.

The last step is actually (-2)×4+5×3/5×4.

In general, if a/b and c/d are two rational numbers then

===================================

《a/c+c/d=a×d+b×c/b×d=ad+bc/nd.》

================================

TO FIND:-

The sum of 5/8 and (-11)/12

SOLUTION:-

hey mate!You could sorten you work as under!

5/8+(-11)/12=5×3+(-11)×2/24=15-22/24

=>(-7)/24.

see below↘⬇↙

1)Divide L.C.M. 24 by denominator 8.The quotient is 3.Multiply 5 by 3.

2)Divide L.C.M.24 by denominator 12.The quotient is 2.Multiply (-11) by 2.

Clarification:-

$\\ \\ \\ \frac{a}{b} + \frac{c}{d} = \frac{a \times \frac{l.c.m}{b} + c \times \frac{l.c.m}{d} }{l.c.m \: of \: b \: and \: d}$

TO FIND:-

The sum of 7/(-18)+5/(-12)+(-9)/(-16)

SOLUTION:-

First write the given rational numbers with positive d enominators.

=>7/(-18)= 7×(-1)/(-18)×(-1)=(-7)/18

=>5/(-12)=5×(-1)/(-12)×(-1)=(-5)/12

=>(-9)/(-16)=(-9)×(-1)/(-16)×(-1)=>9/6

Now,L.C.M.of 18,12&16 is 114.

So,

(-7)/18+(-5)/12+9/16=(-7)×8+(-5×12)+(9×9)/144

=>(-56)-60+81

(-35)/144

Hope it helped!

Answer 2

$\huge \bf Given :-$

$\bf \red{1) By \: taking \: x= \frac {3}{5} \: and \: y = \frac{4}{9}, find \: out \: whether \: (x + y {)}^{ - 1} = {x}^{ - 1} + {y}^{ - 1}</p><p>is \: true \: or \: false.}$

$\bf \huge Solution :-$

$\bf \green {Hence, \: The \: given \: statement \: is \: false.}$

$\huge \bf Given :-$

• $\bf \red{(1) \: \frac{( - 2)}{5} \: and \: \frac{3}{4} }$
• $\bf \red{(2) \: \frac{5}{8} \: and \: \frac{( - 11)}{12} }$
• $\bf \red{(3) \: \frac{7}{(-18)}+ \frac{5}{(-12)}+ \frac {(-9)}{(-16)}}$

$\bf \huge Solution :-$

• $\bf \red{(1) \: \frac{( - 2)}{5} \: and \: \frac{3}{4} =}\bf \green{0.35}$
• $\bf \red{(2) \: \frac{5}{8} \: and \: \frac{( - 11)}{12} =}\bf \green{-0.29166666666}$
• $\bf \red{(3) \: \frac{7}{(-18)}+ \frac{5}{(-12)}+ \frac {(-9)}{(-16)}=}\bf \green{-0.24305555555}$