❖ Solve:
 \sf  \frac{21}{x}  +  \frac{47}{y}  = 110 \\
[tex] \sf \frac{27}{x} + \frac{21}{y} = 162 \\ [/t

Question

❖ Solve:
 \sf  \frac{21}{x}  +  \frac{47}{y}  = 110 \\
 \sf \frac{27}{x}  +  \frac{21}{y}  = 162 \\
❒ Try not to spam !!
❒ Explanation needed !!
❒ I’ill appretiate good responders!!​​

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Eloise 5 months 2021-07-03T07:31:42+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-03T07:32:57+00:00

    Answer:

    21x+47y=110..(1)

    47x+21y=162..(2)

    Multiplying equation (1) by 21, we get

    441x+987y=2310..(3)

    Also multiplying equation (2) by 47, we get

    2209x+987y=7614..(4)

    Now, subtracting equation (3) from (4), we have

    (2209x+987y)−(441x+987y)=7614−2310

    1768x=5304

    x= 1768/5304 =3

    Substituting the value of x in equation (1), we have

    21×3+47y=110

    ⇒47y=110−63

    y= 47/47 =1

    Therefore,

    x=3 and y=1.

    0
    2021-07-03T07:33:29+00:00

    Given equations

     \sf  \frac{21}{x}  + \frac{47v}{y}  = 110 \\  \:  \sf -  -  -  -  -  -  - (i)

     \sf \frac{47}{x}  +  \frac{21}{y}  = 162 \\  \:  \sf -  -  -  -  -  -  - (ii)

     \sf let \: u  =  \frac{1}{x} \:  \: and \:  \: v =  \frac{1}{y}  \\

    So, our equations become

     \sf 21u+ 47v= 110 \:  \:  -  -  -  - (iii)

     \sf 47u+ 21v= 162  \:  \: -  -  -  - (iv)

    From (iii)

     \sf 21u+ 47v =110

     \sf 21u  = 110-47v

     \sf u =  \frac{110}{21}  -  \frac{47v}{21}  \\

    Putting value of u in (iv)

     \sf 47u+ 21v= 162

     \sf \longrightarrow 47( \frac{110}{21}  -  \frac{47v}{21} ) + 21v = 162 \\

    \sf \longrightarrow 47 \times  \frac{110}{21}  - 47 \times  \frac{47v}{21}  + 21v = 162 \\

    \sf \longrightarrow  - 47 \times  \frac{47v}{21}  + 21v = 162 - 47 \times  \frac{110}{21}  \\

    \sf \longrightarrow  \frac{ - 47 \times 47v + 21 \times 21v}{21}  =  \frac{162 \times 21 - 47 \times 110}{21}  \\

    \sf \longrightarrow  \frac{ { - 47}^{2}v +  {21}^{2} v }{21}  =  \frac{162 \times 21 - 47 \times 110}{21}  \\

    \sf \longrightarrow  \frac{ {21}^{2}v -  {47}^{2}v  }{21}  =  \frac{162 \times 21 - 47 \times 110}{21}  \\

    \sf \longrightarrow  {21}^{2} v -  {47}^{2} v = 162 \times 21 - 47 \times 110

    \sf \longrightarrow v( {21}^{2}  -  {47}^{2} ) = 162 \times 21 - 47 \times 110

     \sf \longrightarrow v(21-47) (21+47)  = 162  \times  21-47  \times  110

     \sf \longrightarrow v \times ( - 26) \times 68 = 162 \times 21 - 47 \times 110

    \sf \longrightarrow v \times ( - 26) \times 68 = 3402 - 5170

    \sf \longrightarrow v \times ( - 26) \times 68 =  - 1768

    \sf \longrightarrow v \times 26 \times 68 = 1768

    \sf \longrightarrow v =  \frac{1768}{26 \times 68} \\

    \sf \longrightarrow v =  \frac{221}{13 \times 17}  \\

    \sf \longrightarrow v =  \cancel \frac{7}{7}  \\

    \sf \longrightarrow v = 1

    Putting v = 1 in equation (iii)

     \sf 21u+ 47v= 110

     \sf \longrightarrow 21u+ 47  \times 1= 110

    \sf \longrightarrow 21u + 47 = 110

    \sf \longrightarrow 21u = 110 - 47

    \sf \longrightarrow 21u = 63

    \sf \longrightarrow u =   \cancel\frac{63}{21}  \\

     \sf \longrightarrow u = 3

    Now,

     \sf x =  \frac{1}{u}  =  \frac{1}{3}  \\

     \sf y =   \cancel\frac{1}{1}  = 1 \\

    The answer is

     \sf \: So, x =  \bold{ \frac{1}{3} },  \\  \\  \sf \bold{y = 1} \\

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