According to the question,
[tex]x +y=30~~~~......equation (1)\\xy=176~~~~........equation (2)\\\\Taking~equation~(2),\\ xy=176\\\\y = \frac{176}{x}\\\\Now, ~substuting~the~value~of~y~in~equation~(1), ~we~get,\\\\x+ \frac{176}{x}=30\\\\ \frac{x^2+176}{x}=30\\\\x^2+176=30x\\\\x^2-30x+176=0\\\\x^2-8x-22x+176=0\\\\x(x-8)-22(x-8)=0\\\\(x-8)(x-22)=0\\\\Using~zero~product~property,\\\\Either, \\x-8=0\\x=8\\\\Or,\\ x-22=0 \\x=22\\\\When,\\x=8,\\\\(8)+y=30\\y=22 \\\\And~when,\\x=22\\\\(22)+y=30\\y=8\\ \therefore The~two~numbers~are~8~and ~22[/tex]
