Find three consecutive integers such three times the third integer added to one half of the first integer is 14 less than twice the second integer.

x = first

x + 1 = second {consecutive integers increase by 1 each time}

x + 2 = third

3(x + 2) + 0.5x = 2(x + 1) - 14 {three times 3rd added to half the 1st is 14 less than twice the 2nd}

3x + 6 + 0.5x = 2x + 2 - 14 {used distributive property}

3.5x + 6 = 2x - 12 {combined like terms}

1.5x = -18 {subtracted 2x and 6 from each side}

x = -12 {divided each side by 1.5}

x + 1 = -11 {substituted -12, in for x, into x + 1}

x + 2 = -10 {substituted -12, in for x, into x + 2}

x + 1 = second {consecutive integers increase by 1 each time}

x + 2 = third

3(x + 2) + 0.5x = 2(x + 1) - 14 {three times 3rd added to half the 1st is 14 less than twice the 2nd}

3x + 6 + 0.5x = 2x + 2 - 14 {used distributive property}

3.5x + 6 = 2x - 12 {combined like terms}

1.5x = -18 {subtracted 2x and 6 from each side}

x = -12 {divided each side by 1.5}

x + 1 = -11 {substituted -12, in for x, into x + 1}

x + 2 = -10 {substituted -12, in for x, into x + 2}

**-12, -11, and -10**are the three consecutive integers*- Algebra House*