SOLUTION: Could you please help me with the following problem: The sum of three positive integers is 68. The second is one more than twice the first. The third is three less than the squa

Question 87922This question is from textbook Introductory and Intermediate Algebra
: Could you please help me with the following problem: The sum of three positive integers is 68. The second is one more than twice the first. The third is three less than the square of the first. Find the integers.
I've spent a lot of time on this one with no resolution. The only thing I can come up with is x+y+z=68, y=2x+1, and z=x^2-3. Thank you very much for any assistance you can provide. This question is from textbook Introductory and Intermediate Algebra

Answer by SandySharma(35) (Show Source): You can put this solution on YOUR website!
Let the first integer be x.
Second integer will be 2x + 1.
Third integer will be
It is given that sum of three positive integers is 68.

so,

After simplification we will get

Factorize it, we will get
(x+10)(x-7) = 0
x = -10, 7
As the integer is positive so will take the positive value
Therefore the first integer will be 7
Second integer = 2x + 1 = 2*7 + 1 = 15
Third integer =

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