Question

21x-50y=60,28x-27y=199 find x and y​

Answer 1

Step-by-step explanation:

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Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{Given,}} \\ \\ \\ \: \: \: \: \sf{ 21x - 50y = 60 \: \: - - - - - > (1)} \\ \\ \: \: \: \: \sf{28x - 27y = 199 \: \: - - - - - > (2)}$

$\underline{ \mathfrak{Now,}} \\ \\ \bold{(1) \implies \: - \: 50y = 60 - 21x }\\ \\ \: \: \: \: \: \: \: \bold{ \implies \: 50y = 21x - 60} \\ \\ \: \: \: \: \: \: \: \bold{ \implies \: \red{y= \frac{21x - 60}{50}} \: \: - - - - - > (3)}$

$\underline{ \text{ \: \: Putting the value of \: \red{y} \: in eq. (2) \: \: }}$

$\tt{(2) \implies \:28x - 27( \red{\frac{21x - 60}{50}} ) = 199 }\\ \\ \: \: \: \: \: \: \tt{\implies \: \frac{1400x - (567x - 1620)}{50} = 199} \\ \\ \: \: \: \: \: \: \tt{\implies \: \frac{1400x - 567x + 1620}{50} = 199 }\\ \\ \: \: \: \: \: \tt{ \implies \: \frac{833x + 1620}{50} = 199} \\ \\ \: \: \: \: \: \: \tt{ \implies \:833x + 1620 = 9950 }\\ \\ \: \: \: \: \: \tt{\implies \:833x = 9950 - 1620 }\\ \\ \: \: \: \: \: \: \tt{ \implies \:833x = 8330} \\ \\ \: \: \: \: \: \: \tt{ \implies \:x = \frac{8330}{833} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt{\therefore \: \: \red{x = 10}}$

$\underline{ \text{ \: \: Putting the value of \: \red{x} \: in eq. (2) \: \: }}$

$\tt{(3) \implies \:y = \frac{21 \times \red{10} - 60}{50} } \\ \\ \: \: \: \: \: \: \tt{ \implies \:y = \frac{210 - 60}{50}} \\ \\ \: \: \: \: \: \: \tt{\implies \:y = \frac{150}{50} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\therefore \: \: \red{ y = 3}}$

$\sf{ \huge{ \underline{ \pink{ \: \: x = 10}} \: \: and \: \: \underline{\pink{y = 3}}}}$