Respuesta :
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.
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 [tex]x[/tex] and [tex]y[/tex] are the two integers. Then,
[tex]y=2x-1[/tex] ...(i)
[tex]x^2+y^2=106[/tex] ...(ii)
Using (i) and (ii), we get
[tex]x^2+(2x-1)^2=106[/tex]
[tex]x^2+4x^2-4x+1-106=0[/tex]
[tex]5x^2-4x-105=0[/tex]
Split the middle term.
[tex]5x^2-25x+21x-105=0[/tex]
[tex]5x(x-5)+21(x-5)=0[/tex]
[tex](5x+21)(x-5)=0[/tex]
Using the zero product property, we get
[tex]5x+21=0[/tex]
[tex]x=-\dfrac{21}{5}[/tex]
It is not an integer.
[tex]x-5=0[/tex]
[tex]x=5[/tex]
Substituting [tex]x=5[/tex] in (i), we get
[tex]y=2(5)-1[/tex]
[tex]y=10-1[/tex]
[tex]y=9[/tex]
Therefore, the two integers are 5 and 9.
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