❖ Solve:[tex] \sf \frac{21}{x} + \frac{47}{y} = 110 \\ [/tex][tex] \sf \frac{27}{x} + \frac{21}{y} = 162 \\ [/tex]❒ Try not to spam !!❒ Explanation needed !!❒ I’ill appretiate good responders!! About the author Eloise
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. Reply
Given equations [tex] \sf \frac{21}{x} + \frac{47v}{y} = 110 \\ \: \sf – – – – – – – (i)[/tex] [tex] \sf \frac{47}{x} + \frac{21}{y} = 162 \\ \: \sf – – – – – – – (ii)[/tex] [tex] \sf let \: u = \frac{1}{x} \: \: and \: \: v = \frac{1}{y} \\ [/tex] So, our equations become [tex] \sf 21u+ 47v= 110 \: \: – – – – (iii)[/tex] [tex] \sf 47u+ 21v= 162 \: \: – – – – (iv)[/tex] From (iii) [tex] \sf 21u+ 47v =110[/tex] [tex] \sf 21u = 110-47v[/tex] [tex] \sf u = \frac{110}{21} – \frac{47v}{21} \\ [/tex] Putting value of u in (iv) [tex] \sf 47u+ 21v= 162[/tex] [tex] \sf \longrightarrow 47( \frac{110}{21} – \frac{47v}{21} ) + 21v = 162 \\ [/tex] [tex]\sf \longrightarrow 47 \times \frac{110}{21} – 47 \times \frac{47v}{21} + 21v = 162 \\ [/tex] [tex]\sf \longrightarrow – 47 \times \frac{47v}{21} + 21v = 162 – 47 \times \frac{110}{21} \\ [/tex] [tex]\sf \longrightarrow \frac{ – 47 \times 47v + 21 \times 21v}{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex] [tex]\sf \longrightarrow \frac{ { – 47}^{2}v + {21}^{2} v }{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex] [tex]\sf \longrightarrow \frac{ {21}^{2}v – {47}^{2}v }{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex] [tex]\sf \longrightarrow {21}^{2} v – {47}^{2} v = 162 \times 21 – 47 \times 110[/tex] [tex]\sf \longrightarrow v( {21}^{2} – {47}^{2} ) = 162 \times 21 – 47 \times 110[/tex] [tex] \sf \longrightarrow v(21-47) (21+47) = 162 \times 21-47 \times 110[/tex] [tex] \sf \longrightarrow v \times ( – 26) \times 68 = 162 \times 21 – 47 \times 110[/tex] [tex]\sf \longrightarrow v \times ( – 26) \times 68 = 3402 – 5170[/tex] [tex]\sf \longrightarrow v \times ( – 26) \times 68 = – 1768[/tex] [tex]\sf \longrightarrow v \times 26 \times 68 = 1768[/tex] [tex]\sf \longrightarrow v = \frac{1768}{26 \times 68} \\ [/tex] [tex]\sf \longrightarrow v = \frac{221}{13 \times 17} \\ [/tex] [tex]\sf \longrightarrow v = \cancel \frac{7}{7} \\ [/tex] [tex]\sf \longrightarrow v = 1[/tex] Putting v = 1 in equation (iii) [tex] \sf 21u+ 47v= 110[/tex] [tex] \sf \longrightarrow 21u+ 47 \times 1= 110[/tex] [tex]\sf \longrightarrow 21u + 47 = 110[/tex] [tex]\sf \longrightarrow 21u = 110 – 47[/tex] [tex]\sf \longrightarrow 21u = 63[/tex] [tex]\sf \longrightarrow u = \cancel\frac{63}{21} \\ [/tex] [tex] \sf \longrightarrow u = 3[/tex] Now, [tex] \sf x = \frac{1}{u} = \frac{1}{3} \\ [/tex] [tex] \sf y = \cancel\frac{1}{1} = 1 \\ [/tex] The answer is [tex] \sf \: So, x = \bold{ \frac{1}{3} }, \\ \\ \sf \bold{y = 1} \\ [/tex] ㅤ ㅤ ㅤㅤㅤ be brainly~ Reply
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.
Given equations
[tex] \sf \frac{21}{x} + \frac{47v}{y} = 110 \\ \: \sf – – – – – – – (i)[/tex]
[tex] \sf \frac{47}{x} + \frac{21}{y} = 162 \\ \: \sf – – – – – – – (ii)[/tex]
[tex] \sf let \: u = \frac{1}{x} \: \: and \: \: v = \frac{1}{y} \\ [/tex]
So, our equations become
[tex] \sf 21u+ 47v= 110 \: \: – – – – (iii)[/tex]
[tex] \sf 47u+ 21v= 162 \: \: – – – – (iv)[/tex]
From (iii)
[tex] \sf 21u+ 47v =110[/tex]
[tex] \sf 21u = 110-47v[/tex]
[tex] \sf u = \frac{110}{21} – \frac{47v}{21} \\ [/tex]
Putting value of u in (iv)
[tex] \sf 47u+ 21v= 162[/tex]
[tex] \sf \longrightarrow 47( \frac{110}{21} – \frac{47v}{21} ) + 21v = 162 \\ [/tex]
[tex]\sf \longrightarrow 47 \times \frac{110}{21} – 47 \times \frac{47v}{21} + 21v = 162 \\ [/tex]
[tex]\sf \longrightarrow – 47 \times \frac{47v}{21} + 21v = 162 – 47 \times \frac{110}{21} \\ [/tex]
[tex]\sf \longrightarrow \frac{ – 47 \times 47v + 21 \times 21v}{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex]
[tex]\sf \longrightarrow \frac{ { – 47}^{2}v + {21}^{2} v }{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex]
[tex]\sf \longrightarrow \frac{ {21}^{2}v – {47}^{2}v }{21} = \frac{162 \times 21 – 47 \times 110}{21} \\ [/tex]
[tex]\sf \longrightarrow {21}^{2} v – {47}^{2} v = 162 \times 21 – 47 \times 110[/tex]
[tex]\sf \longrightarrow v( {21}^{2} – {47}^{2} ) = 162 \times 21 – 47 \times 110[/tex]
[tex] \sf \longrightarrow v(21-47) (21+47) = 162 \times 21-47 \times 110[/tex]
[tex] \sf \longrightarrow v \times ( – 26) \times 68 = 162 \times 21 – 47 \times 110[/tex]
[tex]\sf \longrightarrow v \times ( – 26) \times 68 = 3402 – 5170[/tex]
[tex]\sf \longrightarrow v \times ( – 26) \times 68 = – 1768[/tex]
[tex]\sf \longrightarrow v \times 26 \times 68 = 1768[/tex]
[tex]\sf \longrightarrow v = \frac{1768}{26 \times 68} \\ [/tex]
[tex]\sf \longrightarrow v = \frac{221}{13 \times 17} \\ [/tex]
[tex]\sf \longrightarrow v = \cancel \frac{7}{7} \\ [/tex]
[tex]\sf \longrightarrow v = 1[/tex]
Putting v = 1 in equation (iii)
[tex] \sf 21u+ 47v= 110[/tex]
[tex] \sf \longrightarrow 21u+ 47 \times 1= 110[/tex]
[tex]\sf \longrightarrow 21u + 47 = 110[/tex]
[tex]\sf \longrightarrow 21u = 110 – 47[/tex]
[tex]\sf \longrightarrow 21u = 63[/tex]
[tex]\sf \longrightarrow u = \cancel\frac{63}{21} \\ [/tex]
[tex] \sf \longrightarrow u = 3[/tex]
Now,
[tex] \sf x = \frac{1}{u} = \frac{1}{3} \\ [/tex]
[tex] \sf y = \cancel\frac{1}{1} = 1 \\ [/tex]
The answer is
[tex] \sf \: So, x = \bold{ \frac{1}{3} }, \\ \\ \sf \bold{y = 1} \\ [/tex]
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