Verified Answer :

morgeron

they give us 2 pieces to the puzzle.  Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106.  (x^2 + y^2 = 106).

 substitute the value for x into the second equation.

(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106  (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106  (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0  (factor)
(y-5) (5y-21) = 0  (set to 0)
y - 5 = 0
y = 5 

substitute the 5 into the equation for y   (x = 2(5) - 1)
           x = 9   if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.

our answers are 5 and 9.

The two integers are 5 and 9.

Given:

  • One positive integer is 1 less than twice another.
  • The sum of squares of two integers is 106

To find:

The values of integers.

Explanation:

Let \(x\) and \(y\) are the two integers. Then,

\(y=2x-1\)               ...(i)

\(x^2+y^2=106\)       ...(ii)

Using (i) and (ii), we get

\(x^2+(2x-1)^2=106\)

\(x^2+4x^2-4x+1-106=0\)

\(5x^2-4x-105=0\)

Split the middle term.

\(5x^2-25x+21x-105=0\)

\(5x(x-5)+21(x-5)=0\)

\((5x+21)(x-5)=0\)

Using the zero product property, we get

\(5x+21=0\)

\(x=-\dfrac{21}{5}\)

It is not an integer.

\(x-5=0\)

\(x=5\)

Substituting \(x=5\) in (i), we get

\(y=2(5)-1\)

\(y=10-1\)

\(y=9\)

Therefore, the two integers are 5 and 9.

Learn more:

https://brainly.com/question/2463360

Related Questions