Question

find three concecutive even integers such that 3 times the sum of the first two is 48 less than the third

Answer 1

Let x be the smallest integer required, the next two would be (x+2), (x+4)
Since 3 times the sum of the first two is 48 less than the third.
3[ x + (x+2) ] =( x+4 ) - 48
3( 2x +2) = x - 44
6x + 6 = x- 44
Can you solve it from here?

Answer 2

im sorry im still a bit confused

Answer 3

Which step?

Answer 4

ok what is the equasion for the first step?

Answer 5

3(3) times (x) sum of the first two number [ x + (x+2) ] is (=) 48 less than (-48) the third (x+4)
So, you can write it into an equation as
3 x [ x + (x+2)] = (x+4)-48
^ ( times, not unknown x)
Got this step?

Answer 6

yes

Answer 7

And do you understand the rest of the steps?

Answer 8

i think so

Answer 9

That's good :)
Can you work out the answer now?

Answer 10

i let u know in a few min.:)

Answer 11

Okay :)

Answer 12

ok so i got 4x=-46:(

Answer 13

o.o ....
3[ x + (x+2) ] =( x+4 ) - 48
3( 2x +2) = x - 44
6x + 6 = x- 44
6x -x +6 = x-x -44
5x +6 = -44
5x +6-6 = -44-6
5x = -50 ....