Question

One positive integer is 3 less than twice another. the sum of their squares is 170

Answer 1

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