Math, asked by ladekinjal04, 8 months ago

Two numbers differ by 3. The sum of greater number and twice the smaller number is 15. Find the smaller number?​

Answers

Answered by penpencil
3

Answer:

4

Step-by-step explanation:

  • let the two numbers be x and x+3 where x is the smaller number
  • it is given that he sum of greater number and twice the smaller number is 15

so x +3+2x= 15

3x +3=15

3x=15-3=12

x =12/3=4

therefore, the smaller number is 4

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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