Math, asked by tanayadagale23, 11 months ago

two number differ by 3.The sum of the greater number and twice the smaller number is 15 find the smaller number

Answers

Answered by abhi4183
10

let number be x , x+3

atq

x+3+2x=15

4x+3=15

4x=12

x=3


abhi4183: please mark as brainliest
tanayadagale23: it can also be done linear equation in two variable
abhi4183: yes it can
tanayadagale23: ok thank
abhi4183: wel.
Answered by TRISHNADEVI
4

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

 \underline{ \mathfrak{ \: Given, \: }} \\  \\  \text{ \red{The difference between the numbers = 3}} \\  \\  \text{ \red{The sum of the greater number and twice }} \\  \text{ \red{the smaller number = 15}} \\  \\  \underline{ \mathfrak{ \:Suppose, \: }} \\  \\   \:  \:  \:  \:  \:  \:  \:  \: \text {\pink{The greater number = x}} \\  \\  \:  \:  \:  \:  \:  \:  \:   \text{ \pink{The smaller number = y}}

 \bold{ \underline{ \:  \: A.T.Q., \:  \: }} \\  \\  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \: \sf{ \blue{x - y = 3 \:  \:  \:  ------> (1)}} \\  \\  \bold{And, } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf{ \blue{x + 2y  = 15  \:  \: -----> (2)}} \\  \\  \underline{ \bold{ \: Now, \: }} \\  \\  \sf{ \blue{ (1) \implies x = 3 + y  -----> (3)}}

 \underline{ \text{ \: Putting the value of  \pink{x} from eq. (3) in eq.(2), we get, \: }} \\  \\  \tt{(2) \implies x + 2y  = 15} \\  \\  \:  \:  \:  \:  \:  \:  \:  \tt{ \implies (3 + y) + 2y = 15} \\  \\ \:  \:  \:  \:  \:  \:  \:  \tt{ \implies 3 + y + 2y = 15} \\  \\   \:  \:  \:  \:  \:  \:  \:\tt{ \implies 3 + 3y = 15  } \\  \\   \:  \:  \:  \:  \:  \:  \:\tt{\implies 3y = 15 - 3 } \\  \\  \:  \:  \:  \:  \:  \:  \: \tt{\implies 3y = 12} \\  \\  \:  \:  \:  \:  \:  \:  \: \tt{ \implies y = \frac{12}{3} } \\  \\  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \: \tt{\therefore \:  \:  \pink{ y = 4}} \\  \\  \therefore  \:  \:  \text{ \pink{The smallest number, y =  \underline {\: 4  \:} }}

 \underline{\underline{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \: }}\\  \\  \underline{ \mathfrak{ \:  \: Again, \:  \: }} \\  \\   \underline{ \text{ \: Putting the value of  \pink{y} in eq. (3), we get, \: }} \\ \\   \tt{(3) \implies x = 3 + y } \\  \\  \:  \:  \:  \:  \:  \:  \:  \tt{\implies x = 3 + 4 } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{\therefore \:  \:  \pink{ x = 7}} \\  \\ \therefore \:  \:  \text{ \pink{The greater number =  \underline {\: 7 \: }}}

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