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.
Please solve
If anyone solves this question I will mark him/her as brainliest

Answer 1

Answer:

7, 8,9

Step-by-step explanation:

let us consider "y" as the variable

(y X 2) + [(y+1 ) X 3)] + [(y+2) X 4] =74

2y + 3y+3 + 4y + 8 =74

9y + 11 = 74

9y = 74-11

9y = 63

y = 63/9

y = 7

y=7

7+1=8

7+2=9

Answer 2

$\bf \underline{ \underline{\maltese\:Given} }$

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.

$\bf \underline{\underline{\maltese\: To \: find }}$

$\sf \implies Three \: consecutive \: numbers = \: ?$

$\bf \underline{\underline{\maltese\: Solution }}$

$\sf \implies Let \: the \: first \: integer \: be \: x$

$\sf \implies Then, Second \: integer = (x+1)$

$\sf \implies Third \: integer = (x+2)$

$\bf \underline{According \: to \: question},$

$\sf Our \: equation \: will \: be :$

$\bf\implies x \times 2 + 3(x + 1) + 4(x + 2) = 74$

$\sf\implies 2x + 3x + 3+ 4x + 8 = 74$

$\sf Taking \: the \: like \: terms,$

$\sf\implies 2x + 3x + 4x + 3+ 8 = 74$

$\sf\implies 9x + 3+ 8 = 74$

$\sf\implies 9x + 11 = 74$

$\sf\implies 9x = 74 - 11$

$\sf\implies 9x = 63$

$\sf\implies x = \dfrac{63}{9}$

$\sf\implies x = 7$

$\sf Now, these \: numbers \: are :$

$\sf\implies First \: number = \bf x = 7$

$\sf\implies Second \: number = \bf (x+1) = (7+1) = 8$

$\sf\implies Third \: number = \bf (x+2) = (7+2) = 9$

━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\bf \underline{\underline{\maltese\: Verification }}$

$\bf\implies x \times 2 + 3(x + 1) + 4(x + 2) = 74$

$\sf\implies 7 \times 2 + 3 \times 8 + 4 \times 9$

$\sf\implies 14+ 24+ 36$

$\sf\implies 74$

$\bf Hence \: verified \: !$