Find three consecutive integers, such that twice the first added to the last is 23.

x = first integer

x + 1 = second integer

x + 2 = 3rd integer

2x + x + 2 = 23 {twice the first added to the last is 23}

3x + 2 = 23 {combined like terms}

3x = 21 {subtracted 2 from each side}

x = 7 {divided each side by 3}

x + 1 = 8 {substituted 7 for x}

x + 2 = 9 {substituted 7 for x}

x + 1 = second integer

x + 2 = 3rd integer

2x + x + 2 = 23 {twice the first added to the last is 23}

3x + 2 = 23 {combined like terms}

3x = 21 {subtracted 2 from each side}

x = 7 {divided each side by 3}

x + 1 = 8 {substituted 7 for x}

x + 2 = 9 {substituted 7 for x}

**7,8, and 9 are the three consecutive integers***- Algebra House*