please help?

Question

find two consecutive such that the sum of 2times the first integer and 7 times second integer is 88

Answer 1

The two integers are x and x+1

2x + 7(x + 1) = 88
2x + 7x + 7 = 88
9x + 7 = 88
9x = 81
x = 9

So, the integers are 9 and 10

Answer 2

Let the first Integer be x; then the second would be x+1

As per info,
2 (x) + 7 (x+1) = 88
=> 2x+7x+7=88
=> 9x+7=88
=> 9x=81
=> x=9

So the two numbers are 9 and 10

Answer 3

2 x 2 + 7 x 12 = 88

Answer 4

Because they are consecutive numbers,

Let X= the first integer
Let Y= the second integer

Y= (X + 1) (y comes after x by one number - it's consecutive)

Thus we write the equation:

2x + 7y = 88
substitute for y
2x + 7(x+1) = 88 (then multiply through)
2x + 7x + 7 = 88 ( then combine like terms, in this case, x)
9x + 7 = 88 (then minus the non-unknown to other side, in this case, -7 to both sides)
9x = 81 (then divide by factor of x, in this case, 9)
x = 9

if x=9 and y= x + 1
then y = 10

plug in values to check answer:

2(9) + 7(10) = 88

which it does

x = 9, y = 10

Answer 5

n = first number
n+1=second number

2n+7(n+1)=88
2n+7n+7=88
9n=81
n = 9
n+1 = 10

9 and 10