Respuesta :

chi3fcomm5nd
let x=number1 and y=number2
[tex] {x}^{2} + {y}^{2} = 170 \\ 2x - 3 = y \\ {x}^{2} + {(2x - 3)}^{2} = 170 \\ {x}^{2} + 4 {x}^{2} - 12x + 9 = 170 \\ 5 {x}^{2} - 12x + 9 - 170 = 0 \\ 5 {x}^{2} - 12x - 161 = 0 \\ (x - 7)(5x + 23) = 0 \\ x = 7 \: or \: x = \frac{ - 23}{5} [/tex]
x is postive the x must be =7
[tex]y = 2x - 3 \\ y = 2(7) - 3 \\ y = 14 - 3 \\ y = 11[/tex]
checking our soultion
[tex] {x}^{2} + {y}^{2} = 121 \\ {7}^{2} + {11}^{2} = 49 + 121 = 170[/tex]
which makes our soultion correct

The two positive integers in which one positive integer is 3 less than twice another, and the sum of their squares is 170 are 7 and 11

Let the first positive integer = x

Let the other positive integer = y

The first positive integer is 3 less than twice the other

x  =  2y  -  3.......(1)

The sum of the squares of the two positive integers is 170

x²  +  y²  =  170........(2)

Substitute equation (1) into equation (2)

(2y - 3)²  +y²  =  170

4y²  -  12y  +  9  +  y²  =  170

Collect like terms

4y²  +  y²  -  12y  +  9  -  170  =  0

5y²  -  12y  -  161   =   0

Solving the quadratic equation above

(5y + 23)(y - 7)   =  0

y - 7  =  0

y   =  7

5y   +  23   =  0

5y   =  -23

y   =  -23/5

Since the number has to be a positive integer, y = 7

Substitute y = 7 into the equation x = 2y - 3

x   =  2(7)  -  3

x   =  14  -  3

x   =  11

Therefore, the two positive integers are 7 and 11

Learn more here: https://brainly.com/question/6063476

Otras preguntas