Question

three consecutive integers are such that when they are taken in increasing order and

multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.​

Answer 1

Answer:

Let us consider the number to be

First number = x

Second Number = x+1

Third Number = x+2

According to the question, We get

2x + 3(x+1) + 4(x+2)-74

2x+ 3x+3+4x+8=74

9x + 11=74

9x = 74-11

9x = 63

x = 63/9

x= 7

Therefore,

The First Number = x = 7

The second number = x+1 = 7+1 = 8

the third Number = x+2 = 7+2 = 9

Hence The first,Second & third Number are 7,8 & 9 respectively

Pls Mark me the brainliest

Answer 2

Given : three consecutive integers are such that when they, are taken in increasing order and multiplied by 2,3 and 4 respectively, they add up to 74.

⠀⠀⠀⠀⠀To Find : The numbers respectively

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀

❒ Let the numbers be x, x + 1, x + 2 respectively

${ \underline{ \bf{ \bigstar \: According \: to \: the \: question : }}}$

• When the numbers are multiplied with the numbers 2,3,4 respectively they add up to 74

${ \underline{ \frak{As \: we \: know \: that \dag}}}$

• The numbers are x , x + 1 and x + 2 so, now let's frame an equation according to the question

${ \underline{ \bf{ \star \: Framing \: an \: equation \: we \: get : }}}$

${ : \implies} \sf \: (x \times 2) + \{(x + 1) \times 3 \} \: +\{(x + 2) \times 4 \} = 74 \\ \\ \\ { : \implies} \sf 2x + 3x + 3 + 4x + 8 = 74 \\ \\ \\ { : \implies} \sf 9x + 11 = 74 \\ \\ \\ { : \implies} \sf 9x = 74 - 11 \\ \\ \\ { : \implies} \sf 9x = 63 \\ \\ \\ { : \implies} \sf x = \cancel \frac{63}{9} \\ \\ \\ { \dashrightarrow} \sf { \pink{ \underline{ \boxed{ \frak{x = 7}}} \star}}$

Hence The numbers are,

⠀⠀⠀

⠀⠀⠀⠀☆ x = (7) = 7

⠀⠀⠀⠀☆ x + 1 = ( 7 + 1 ) = 8

⠀⠀⠀⠀☆ x + 2 = ( 7 + 2 ) = 9

${ \underline{ \rm{ \therefore the \: numbers \: are \: 7 , 8 \: and \: 9 \: respectively}}}$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀