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Solve the system of equations.\newliney=x2−11x+27y = x^2 - 11x + 27\newliney=−22x+3y = -22x + 3\newline\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=x2−11x+27y = x^2 - 11x + 27\newliney=−22x+3y = -22x + 3\newline\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)\newline(______,______)
  1. Set Equations Equal: We have the system of equations:\newliney=x2−11x+27y = x^2 - 11x + 27\newliney=−22x+3y = -22x + 3\newlineTo find the solution, we will set the two equations equal to each other because they both equal yy.\newlinex2−11x+27=−22x+3x^2 - 11x + 27 = -22x + 3
  2. Rearrange and Standardize: Now we will rearrange the equation to bring all terms to one side and set it equal to zero, which will give us a standard form of a quadratic equation.\newlinex2−11x+27+22x−3=0x^2 - 11x + 27 + 22x - 3 = 0\newlineCombining like terms, we get:\newlinex2+11x+24=0x^2 + 11x + 24 = 0
  3. Factor Quadratic Equation: Next, we need to factor the quadratic equation. We are looking for two numbers that multiply to 2424 and add up to 1111. These numbers are 88 and 33. So we can write the factored form as: \newline(x+8)(x+3)=0(x + 8)(x + 3) = 0
  4. Solve for x: Now we will solve for x by setting each factor equal to zero.\newlineFirst, for (x+8)=0(x + 8) = 0, we get x=−8x = -8.\newlineSecond, for (x+3)=0(x + 3) = 0, we get x=−3x = -3.
  5. Find y-values: With the x-values found, we now need to find the corresponding y-values. We can substitute xx back into either of the original equations. We'll use the second equation y=−22x+3y = -22x + 3 for simplicity.\newlineFor x=−8x = -8, we substitute into the equation to get y=−22(−8)+3=176+3=179y = -22(-8) + 3 = 176 + 3 = 179.\newlineFor x=−3x = -3, we substitute into the equation to get y=−22(−3)+3=66+3=69y = -22(-3) + 3 = 66 + 3 = 69.

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