Question

SOLVE THE EQUATION-

21\x +47\y =110
27\x + 21\y=162 x and y is not equal to zero

give me proper solution please​

Answer 1

Answer

21x+47y=1101)

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

Multiplying equation (1) by 21, we get

441x+987y=23103)

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.

Answer 2

$\sf \dfrac{21}{x} + \dfrac{47}{y} = 110 \:$

$\sf \dfrac{47}{x} + \dfrac{21}{y} = 162$

Let,

$\sf \dfrac{1}{x} = a \: \: \: and \: \: \: \dfrac{1}{y} = b$

$\sf\therefore~~ 21a+47b=110 ~~~~~~----(1) \: \: \: \: \: \: \: ~~ \\ \sf 47a+21b=162~~~~~~----(2) \: \: \:$

• On Solving Equation (1)

$\sf \: \: \: \implies \: 21a+47b=110$

$\sf \: \: \: \implies \: 21a=110 - 47b$

$\sf \: \: \: \implies \: a= \dfrac{ 110 - 47b}{21}$

• Substituting the value of a in Equ.(2)

$\sf \: \: \: \implies \: 47 \bigg( \dfrac{110 - 47b}{21} \bigg) + 21b = 162$

$\sf \: \: \: \implies \: \dfrac{5170 - 2209b}{21} + 21b = 162$

$\sf \: \: \: \implies \: \dfrac{5170 - 2209b + 441b}{21} = 162$

$\sf \: \: \: \implies \: \dfrac{5170 - 1786b}{21} = 162$

$\sf \: \: \: \implies \: {5170 - 1786b} = 162 \times 21$

$\sf \: \: \: \implies \: {5170 - 1786b} = 3402$

$\sf \: \: \: \implies \: { - 1786b} = 3402 - 5170$

$\sf \: \: \: \implies \: { \cancel{- 1786}b} = \cancel{ - 1786}$

$\sf \: \: \: \implies \: b = 1$

• $\sf{Putting \: the \: value \: of \: b \: in \: \bf \: a = \dfrac{110 - 47b}{21} }$

${\: \: \: \: \: \implies \sf{a = \dfrac{110 - 47b}{21}}}$

${\: \: \: \: \: \implies \sf{a = \dfrac{110 - 47 \times 1}{21}}}$

${\: \: \: \: \: \implies \sf{a = \dfrac{110 - 47}{21}}}$

${\: \: \: \: \: \implies \sf{a = \dfrac{63}{21}}}$

${\: \: \: \: \: \implies \sf{a = 3}}$

therefore,

$\sf a= \dfrac{1}{x} = 3 \: \: \: \: \: \: \: and \: \: \: \: b = \dfrac{1}{y} = 1$

$\bf x = \dfrac{1}{ 3} \: \: \: \: \: \: \: and \: \: \: \: y= 1$