Find two consecutive integers such that the sum of 7 times the first integer and 4 times second integer is 92.

Question

I can't seem to get these.

Answer 1

let x=first integer, x+1=2nd integer

7x+4(x+1)=92

7x+4x+4=92

11x=88

x=8

8 and 9 are the numbers

Answer 2

Define the following:
x = "the first integer"
x + 1 = "the second integer"

7x + 4(x + 1) = 92
7x + 4x + 4 = 92
11x + 4 = 92
11x = 88
x = 8

Therefore, the integers are 8 and 9.

Answer 3

First integer = x
Consecutive integer = x+1

7*x+4*(x+1) = 92

7x+4x+4=92
11x=88
x=8
x+1=9

Answer 4

let x=first interger x+1=2nd interger

7x+4(x+1)=92 use the distributive property on 2nd inter.
7x+4x=4=92 Now add -4 from both sides
7x+4x+4-4=92-4
7x+4x=88 Add like terms
11x=88 Divide both sides by 11
x=8 for the 1st integer
x+1=9(2nd integer)
Check your work by plugging answers into original problem; 7(8)+4(9)=92
56+36=92
92=92

Answer 5

n = first integer
n + 1 = second integer
7*n + 4 (n + 1) = 92
11 n + 4 = 92
11 n = 92 - 4 = 88
n = 88 / 11 = 8
First integer = n
= 8
Second integer = n + 1
= 8 + 1 + 9

Answer 6

Assuming the smaller integer is x, then the larger one is x+1

7*x + 4*(X+1) = 92
then you solve for X, which would give you X=8

So the two integers would be 8 and 9. =)

Answer 7

x = first consecutive number

7x + 4(x+1) = 92
7x + 4x + 4 = 92
11x = 88
x = 8

hence, 8 and 9

Answer 8

7x + 4(x+1) = 92
7x+4x+4=92
11x=88
x=8

the integers are 8 and 9

to check 7x8 + 4x9 = 56+36 = 92

Answer 9

x = 1st integer
x+1 = 2nd integer
7x +4(x+1)=92
7x +4x +4 = 92
11x = 88
x = 8
x+1 =9

Answer 10

7(8) + 4(9) = 92

8 and 9

Merry Christmas