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Solve the system of equations. {:[y=22 x-38],[y=x^(2)+11 x-14]:} Write the coordinates in exact form. Simplify all fractions and radicals. {:[(-","-)],[(-","-)]:}

Solve the system of equations.\newliney=22x−38y=x2+11x−14\begin{array}{l}y=22 x-38 \\y=x^{2}+11 x-14\end{array}\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(−,−)(−,−)\begin{array}{l}(-,-) \\(-,-)\end{array}

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Q. Solve the system of equations.\newliney=22x−38y=x2+11x−14\begin{array}{l}y=22 x-38 \\y=x^{2}+11 x-14\end{array}\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(−,−)(−,−)\begin{array}{l}(-,-) \\(-,-)\end{array}
  1. Set Equation to Zero: Now we will rearrange the equation to set it to zero and solve for xx.
    x2+11x−14−(22x−38)=0x^2 + 11x - 14 - (22x - 38) = 0
    x2+11x−14−22x+38=0x^2 + 11x - 14 - 22x + 38 = 0
    x2−11x+24=0x^2 - 11x + 24 = 0
  2. Factor Quadratic Equation: Next, we factor the quadratic equation.\newline(x−3)(x−8)=0(x - 3)(x - 8) = 0\newlineThis gives us two possible solutions for xx:\newlinex=3x = 3 or x=8x = 8
  3. Substitute x=3x = 3: Now we will substitute x=3x = 3 into the first equation to find the corresponding yy value.\newliney=22(3)−38y = 22(3) - 38\newliney=66−38y = 66 - 38\newliney=28y = 28\newlineSo one solution is (3,28)(3, 28).
  4. Substitute x=8x = 8: Next, we substitute x=8x = 8 into the first equation to find the corresponding yy value.\newliney=22(8)−38y = 22(8) - 38\newliney=176−38y = 176 - 38\newliney=138y = 138\newlineSo the second solution is (8,138)(8, 138).

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