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Solve the system of equations.\newliney=45x2−6x−20y = 45x^2 - 6x - 20\newliney=−6x+25y = -6x + 25\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)

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Q. Solve the system of equations.\newliney=45x2−6x−20y = 45x^2 - 6x - 20\newliney=−6x+25y = -6x + 25\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)
  1. Set Equations Equal: We have the system of equations:\newliney=45x2−6x−20y = 45x^2 - 6x - 20\newliney=−6x+25y = -6x + 25\newlineTo find the intersection points, we set the two equations equal to each other.\newline45x2−6x−20=−6x+2545x^2 - 6x - 20 = -6x + 25
  2. Simplify Equation: Now, we simplify the equation by moving all terms to one side to set the equation to zero. \newline45x2−6x−20+6x−25=045x^2 - 6x - 20 + 6x - 25 = 0\newline45x2−45=045x^2 - 45 = 0
  3. Combine Like Terms: Next, we simplify the equation further by combining like terms.\newline45x2−45=045x^2 - 45 = 0\newline45(x2−1)=045(x^2 - 1) = 0
  4. Factorize: We recognize that x2−1x^2 - 1 is a difference of squares, which can be factored as (x+1)(x−1)(x + 1)(x - 1). \newline45(x+1)(x−1)=045(x + 1)(x - 1) = 0
  5. Set Factors Equal: To find the values of xx, we set each factor equal to zero.(x+1)=0(x + 1) = 0 or (x−1)=0(x - 1) = 0Solving for xx gives us x=−1x = -1 and x=1x = 1.
  6. Find Y-Values: Now that we have the x-values, we need to find the corresponding y-values by substituting xx back into one of the original equations. We can use y=−6x+25y = -6x + 25 for simplicity.\newlineFor x=−1x = -1: y=−6(−1)+25=6+25=31y = -6(-1) + 25 = 6 + 25 = 31\newlineFor x=1x = 1: y=−6(1)+25=−6+25=19y = -6(1) + 25 = -6 + 25 = 19
  7. Coordinates: We have found the yy-values corresponding to the xx-values. Therefore, the coordinates of the intersection points in exact form are:\newlineFirst Coordinate: (−1,31)(-1, 31)\newlineSecond Coordinate: (1,19)(1, 19)

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