Type the correct answer in each box. Use numerals instead of words.Consider the systems of equations below.

Answer:System A has 4 real solutions.System B has 0 real solutions.System C has 2 real solutionsStep-by-step explanation:System A:x^2 + y^2 = 17   eq(1)y = -1/2x            eq(2)Putting value of y in eq(1)x^2 +(-1/2x)^2 = 17x^2 + 1/4x^2 = 175x^2/4 -17 =0Using quadratic formula:[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]a = 5/4, b =0 and c = -17[tex]x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}[/tex]Finding value of y:y = -1/2x[tex]y=-1/2(\frac{\pm2\sqrt{85}}{5})[/tex][tex]y=\frac{\pm\sqrt{85}}{5}[/tex]System A has 4 real solutions.System By = x^2 -7x + 10    eq(1)y = -6x + 5            eq(2)Putting value of y of eq(2) in eq(1)-6x + 5 = x^2 -7x + 10=> x^2 -7x +6x +10 -5 = 0x^2 -x +5 = 0Using quadratic formula:[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]a= 1, b =-1 and c =5[tex]x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}[/tex]Finding value of y: y = -6x + 5y = -6(\frac{1\pm\sqrt{19}i}{2})+5Since terms containing i are complex numbers, so System B has no real solutions.System B has 0 real solutions.System Cy = -2x^2 + 9    eq(1)8x - y = -17        eq(2)Putting value of y in eq(2)8x - (-2x^2+9) = -178x +2x^2-9 +17 = 02x^2 + 8x + 8 = 02x^2 +4x + 4x + 8 = 02x (x+2) +4 (x+2) = 0(x+2)(2x+4) =0x+2 = 0 and 2x + 4 =0x = -2 and 2x = -4x =-2 and x = -2So, x = -2Now, finding value of y:8x - y = -17    8(-2) - y = -17    -16 -y = -17-y = -17 + 16-y = -1y = 1So, x= -2 and y = 1System C has 2 real solutions