Chapter 8 - Section 8.6 - Systems of Nonlinear Equations - 8.6 Assess Your Understanding - Page 616: 74

$b=\pm2$ and $a=\pm5$

the product of two numbers, $ab=10$ and the difference of their squares, $a^2-b^2=21,$ Thus, $a=\frac{10}{b},$ $(\frac{10}{b})^2-b^2=21,$ $\frac{100}{b^2}-b^2=21,$ $100-b^4=21b^2,$ $b^4+21b^2-100=0,$ let $b^2=k$ $k^2+21k-100=0,$ $k^2-4k+25k-100=0,$ $k(k-4)+25(k-4)=0,$ $(k-4)(k+25)=0,$ Thus, $(b^2+25)(b^2-4)=0,$ $b=\pm2$ and $a=\pm5$