SOLUTION: Find five consecutive intgers if the square of the first less the square of the third is 5 less than the sum of the fourth and the fifth intgers.

Question 325633: Find five consecutive intgers if the square of the first less the square of the third is 5 less than the sum of the fourth and the fifth intgers.
Answer by jessica43(140) (Show Source): You can put this solution on YOUR website!
To solve this problem, we need to write an equation using the information provided in the equation.
First, you have five consecutive integers:
x, x+1, x+2, x+3, x+4 (where x = the first integer)
Second, we know the square of the first less the square of the third is 5 less than the sum of the fourth and the fifth intgers. This can be written as:
x^2 - (x+2)^2 = ((x+3)+(x+4)) - 5
Now we can solve for x:
x^2 - (x+2)^2 = ((x+3)+(x+4)) - 5
x^2 - (x+2)^2 = (2x + 7)- 5
x^2 - (x+2)^2 = 2x + 2
x^2 - (x+2)(x+2) = 2x + 2
x^2-x^2+4x+4 = 2x + 2
4x + 4 = 2x + 2
2x +4 = 2
2x = -2
x = -1
So the first integer in the series is -1. So the five consecutive integers are -1, 0, 1, 2, 3
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