Question

find three consecutive even natural numbers such that 7 times the middle number is 42 less 4 times the some of other 2.

Answer 1

Answer: 2 ; 4 ; 6

Step-by-step explanation:

Let the terms be a-2 , a , a+2

Middle term =a

Sum of other two = (a-2)+(a+2)=a-2+a+2= 2a

ATP ,

7a= 4*2a - 4

7a= 8a-4

7a-8a=-4

-a=-4

a=4

Terms = 2 , 4 ,6

Answer 2

Answer:

40,42 and 44.

Step-by-step explanation: let the consecutive numbers be  x, x+2,x+2+2 (even numbers differ by 2)

the middle number is x+2.

seven times the number is 7(x+2)

=7x+14.

4 times the sum of the other two is:

(x)+(x+2+2)

=2x+4

four times of this is:

4(2x+4)

=8x+16

hence, 7 times the middle number is 42 less than 4 times the sum of the other two

this means that,

(7x+14)+42=8x+16

7x+56=8x+16

minus 7x on both sides.

56=x+16

minus 16 on both sides

40=x

x=40

hence the three consecutive numbers are x, x+2 and x+2+2

so,

40, 40+2 and 40+2+2

which gives us 40,42 and 44.