The sum of twice an integer and 3 times the next consecutive integer is 48

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The sum of twice an integer and 3 times the next consecutive integer is 48

Let the first integer be n
This means the next consecutive integer is n + 1

Twice an integer means we multiply n by 2:
2n

3 times the next consecutive integer means we multiply (n + 1) by 3
3(n + 1)

The sum of these is:
2n + 3(n + 1)

The word is means equal to, so we set 2n + 3(n + 1) equal to 48:
2n + 3(n + 1) = 48

Solve for n in the equation 2n + 3(n + 1) = 48

We first need to simplify the expression removing parentheses

Simplify 3(n + 1): Distribute the 3 to each term in (n+1)
3 * n = (3 * 1)n = 3n
3 * 1 = (3 * 1) = 3
Our Total expanded term is 3n + 3

Our updated term to work with is 2n + 3n + 3 = 48

We first need to simplify the expression removing parentheses
Our updated term to work with is 2n + 3n + 3 = 48

Step 1: Group the n terms on the left hand side:
(2 + 3)n = 5n

Step 2: Form modified equation
5n + 3 = + 48

Step 3: Group constants:
We need to group our constants 3 and 48. To do that, we subtract 3 from both sides
5n + 3 - 3 = 48 - 3

Step 4: Cancel 3 on the left side:
5n = 45

Step 5: Divide each side of the equation by 5
5n/5 = 45/5

Cancel the 5's on the left side and we get:
n = 9