Find the LARGEST of three consecutive integers such that 3 times the sum of the first and the third integer is equal to 30 more than 4 times the second integer.

Answer:
The largest would be 16 (the other two would be 14, 15).

Explanation:
Solve this by:
1- setting the first equal to x, the second to x+1 and the third to x+2.

2- write the equation given the description.
3(x+x+2) = 4(x+1) + 30.

3- solve.
3x + 3x + 6 = 4x + 4 + 30.
6x + 6 = 4x + 34.
2x = 28.
x =14

Answer Link

eurydike eurydike · Answer 2 · 2019-04-26T19:30:15+02:00

Answer:

16

Step-by-step explanation:

Let the three consecutive integers be

X

X+1

X+2

The sum of first and third integer = (X ) + (X+2)

= 2X+2

3 times of sum of first and third integer = 3 (2X+2) -----------A

Four times of second integer = 4 (X+1)

30 more than four times of second integer = 30 + (4 (X+1))…………….B

Equating equation A and B, we get –

3 (2X+2) = 30 + (4 (X+1))

6 X + 6 = 30 + 4 X + 4

2 X = 28

X= 14

X + 1 = 15

X+ 2 = 16

So the largest of the three integer is 16