Find three consecutive integers such that twice the sum of the first two is to less than three times the third

Step-by-step explanation:

Let the consecutive integers be x - 1 , x , x + 1

Equation

2*(x - 1 + x) = 3*(x+1) - 2

Solution

2(2x - 1) = 3(x + 1) - 2           After the first two have been combined remove the brackets.

4x - 2 = 3x + 3 - 2               Combine the right

4x - 2 = 3x + 1

4x - 2 + 2 = 3x + 1 + 2         Add 2 to both sides

4x = 3x + 3                          Subtract 3x from both sides

4x-3x = 3x - 3x + 3             Combine

x = 3

Check

Left hand side

2(2 + 3)

2(5)

10

Right hand side

3*(x+1) - 2

3*(4) - 2

12 - 2

10

The answers at least agree.

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step-by-step explanation: