Answer:The other solution of the system is [tex](x,y) = (\frac{4}{5} ,\frac{3}{5} )[/tex]Step-by-step explanation:Here, the given set of equations are: y = 2x -1 and [tex]x^{2} + y^{2} =1[/tex]Now, to solve them put y = 2x -1 in the second equation, so we get[tex]x^{2} + y^{2} =1[/tex] ⇒ [tex]x^{2} + (2x-1)^{2} =1[/tex]Also, by ALGEBRAIC IDENTITY : [tex](a-b)^{2} = a^{2} + b^{2} -2ab[/tex]So, implying above , we get[tex]x^{2} + (2x-1)^{2} =1[/tex] ⇒ [tex]x^{2} + 4x^{2} + 1 -4x =1 or, 5x^{2} -4x =0[/tex]Now, [tex]5x^{2} -4x =0[/tex] ⇒x(5x-4) =0So, either x =0, or x = 4/5As (0, -1) is already a given solution:So, when x = 4/5 ,put in equation (1)y = 2(5/2) -1 = (3/5)Hence, the other solution of the system is [tex](x,y) = (\frac{4}{5} ,\frac{3}{5} )[/tex]