Question

Find three consecutive even integers such that 27 more than 3 times the second is 7 less than 8 times the first

Answer 1

Answer:

The three consecutive even integers are 8, 10 and 12.

Step-by-step explanation:

Let the three consecutive even integers be $x, x+2, x+4$

The statement "27 more than 3 times the second is 7 less than 8 times the first" can be expressed as:

$27+3(x+2)=8x-7$

Solve this expression for x as follows:

$27+3(x+2)=8x-7\\27+3x+6=8x-7\\33+3x=8x-7\\40=5x\\x=8$

The three consecutive even integers are:

$x=8\\x+2=8+2=10\\x+4=8+4=12$

Answer 2

The three consecutive even integers are 8, 10 and 12.

The calculation is:

Let us assume the three consecutive even integers be x, x +2, and x +4

Since the statement

The statement "27 more than 3 times the second is 7 less than 8 times the first" can be expressed as:

27 + 3(x + 2) = 8x - 7

27 + 3x + 6 = 8x - 7

33 + 3x = 8x - 7

40x = 5x

x = 8

So,

x = 8

x + 2 = 8 + 2 = 10

x + 4 = 8 + 4 = 12

Find out more information about the integer here:https://brainly.com/question/1768254?referrer=searchResults