Math, asked by abbavaramgowthami95, 9 months ago

21x-50y=60,28x-27y find x and y​

Answers

Answered by TRISHNADEVI
1

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: QUESTION \:  \: } \mid}}}}}

 \green{ \text{If \: 21x - 50y =  60 \: and \: 28x  - 27y = 199, find}} \\  \green{ \text{ x and y.}}

 \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}}}}

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