Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Problem Solving - Page 663: 41b

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From part a, we found $y=-\frac{1}{7}x+\frac{5}{7}$ Substitute values of $x$ to get the value of $y$: $y_1=-\frac{1}{7}x_1+\frac{5}{7} \rightarrow y_1=\frac{4}{5}\\ y_1=-\frac{1}{7}x_2+\frac{5}{7} \rightarrow y_2=\frac{3}{5}$ The condition is $x^2+y^2\leq1$ The solutions are $(-\frac{3}{5},\frac{4}{5});(\frac{4}{5},\frac{3}{5})$