Find two consecutive integers such that the sum of 7 times the first integer and 3 times second integer is 103

Question

Find two consecutive integers such that the sum of 7 times the first integer and 3 times second integer is 103?

Two sonsecutuve integers that equal 103 is 51 and 52, I know that part. I have played with this question for way too long. Help?

Answer 1

7x + 3(x + 1) = 103
10x + 3 = 103
10x = 100
x = 10
x + 1 = 11
Check:

7(10) + 3(11) = 103
70 + 33 = 103
103 = 103

Answer 2

Define two integers, X and Y such that Y = X + 1 Then, 2X + 4Y = 46 Substituting: 2X + 4(X + 1) = 46 2X + 4X + 4 = 46 6X + 4 - 4 = 46 - 4 6X = 42 X = 42/6 = 7 Y = X + 1 = 7 + 1 = 8 The integers are 7 and 8.

Answer 3

Let the consecutive integers be x & y
Now,
7x + 3y = 103
7x + 3( x + 1 ) =103
7x + 3x + 3 = 103
10x = 103 - 3
10x = 100
x = 10
Hence, y = x + 1 = 11
Integers are 10 & 11.

Answer 4

The first integer is 10, 10 x 7 = 70
The second is 11, 11x 3 = 33
70 + 33 =103

You prolly neglected to notice they stated that 103 was the SUM of these two integers.

Answer 5

Integers are n, n + 1

the sum of 7 times the first integer and 3 times second integer is 103

7n + 3(n + 1) = 103

7n + 3n + 3 = 103

10n = 100

n = 10

Integers are 10, 11

Answer 6

Using some basic algebra, the question can be expressed as:

7X+3(X+1)=103

Multiply the midsection and you get:

7X+3X+3=103

Subtract 3 from each side of the equation to get:

7X+3X=100

That converts easily to:

10X=100

Divide each side of the equation by 10 to get:

X=10

So the first integer (X) would be 10 and the second (X+1) would be 11.

Answer 7

Let the first number be X
and the second number be X+1

So seven times the first number is
7X
and three times the second number is
3(X+1)

So 7X+3(X+1)=103
distribute the 3
7X+3X+3=103
combine like terms
10X+3=103
Subtract 3 from both sides
10X=100
Divide both sides by 10
X=10

So your first number is 10 and your second number is 10+1 which is 11. Try it out by plugging those values in your original equation

7(10)+3(11)=103

And so it does!

Answer 8

Let the numbers be x and x+1
Then 7x+3(x+1)=103
10x+3=103
x=10
So the two numbers would be 10 and 11.

Answer 9

7x + 3(x + 1) = 103
10x = 100
x = 10
Integers are 10 and 11

Answer 10

x
x+1

7x+3(x+1)=103
7x+3x+3=103
10x=103-3
10x=100
x=50
x+1=50+1=51