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Solve the system of equations.\newliney=x2−40x+20y = x^2 - 40x + 20\newliney=−21x−28y = -21x - 28\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)

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Solve a system of linear and quadratic equations: parabolasright chevron icon

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Q. Solve the system of equations.\newliney=x2−40x+20y = x^2 - 40x + 20\newliney=−21x−28y = -21x - 28\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)
  1. Set Equations Equal: Set the two equations equal to each other since they both equal yy.x2−40x+20=−21x−28x^2 - 40x + 20 = -21x - 28
  2. Move Terms to One Side: Move all terms to one side to set the equation to zero.\newlinex2−40x+21x+20+28=0x^2 - 40x + 21x + 20 + 28 = 0\newlinex2−19x+48=0x^2 - 19x + 48 = 0
  3. Factor Quadratic Equation: Factor the quadratic equation. \newline(x−16)(x−3)=0(x - 16)(x - 3) = 0
  4. Solve for x: Solve for x by setting each factor equal to zero.\newlinex−16=0x - 16 = 0 or x−3=0x - 3 = 0\newlinex=16x = 16 or x=3x = 3
  5. Substitute xx into Second Equation: Substitute x=16x = 16 into the second equation to find yy.\newliney=−21(16)−28y = -21(16) - 28\newliney=−336−28y = -336 - 28\newliney=−364y = -364
  6. Find Coordinates: Substitute x=3x = 3 into the second equation to find yy.\newliney=−21(3)−28y = -21(3) - 28\newliney=−63−28y = -63 - 28\newline$y = \(-91\)
  7. Find Coordinates: Substitute \(x = 3\) into the second equation to find \(y\).\(\newline\)\(y = -21(3) - 28\)\(\newline\)\(y = -63 - 28\)\(\newline\)\(y = -91\)Write the coordinates in exact form.\(\newline\)First Coordinate: \((16, -364)\)\(\newline\)Second Coordinate: \((3, -91)\)

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