Question

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

Answer 1

Step-by-step explanation:

Let the three consecutive integers be x, x+1, x+2 in increasing order

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 74

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=74

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=742x+3x+3+4x+8=74

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=742x+3x+3+4x+8=749x+11=74

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=742x+3x+3+4x+8=749x+11=749x=74-11=63

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=742x+3x+3+4x+8=749x+11=749x=74-11=63X=63÷9=7

Let the three consecutive integers be x, x+1, x+2 in increasing orderMultiply 2,3,4 respectively2(x)+3(x+1)+4(x+2)This aad up to 742(x)+3(x+1)+4(x+2)=742x+3x+3+4x+8=749x+11=749x=74-11=63X=63÷9=7Therefore the three consecutive integers are 7,8,9

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Answer 2

Answer:

Hi!!Mate here is your answer

Refer to the attachment:

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