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valina [46]
3 years ago
10

Find four consecutive even integers such that seven times the first exceeds their sum by 18.

Mathematics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

10, 12, 14, 16

Step-by-step explanation:

Given: Four consecutive even integers such that seven times the first exceeds their sum by 18.

Lets assume the first number be "x".

As it is even integers

∴ Second consecutive number will be (x+2)

  Third consecutive number will be (x+4)

  Fourth consecutive number will be (x+6)

Now, as given seven times the first exceeds their sum by 18.

∴ [x+(x+2)+(x+4)+(x+6)]+18 = 7x

solving the equation to find the number.

⇒ [x+(x+2)+(x+4)+(x+6)]+18 = 7x

Opening parenthesis

x+x+2+x+4+x+6+18 = 7x

⇒ 4x+30= 7x

subtracting both side by 4x

⇒ 30= 3x

Dividing both side by 3

∴ x= 10.

Hence, subtituting the value x to find four consecutive even integers.

  First number is 10

  Second consecutive number will be (10+2) = 12

  Third consecutive number will be (10+4)= 14

  Fourth consecutive number will be (10+6) = 16

∴ Four consecutive even integers are 10,12,14,16.

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If 7x+3=24 what is the value of 5-4x?
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Step-by-step explanation:

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