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Solve the system of equations.\newliney=28x2y = -28x - 2\newliney=x214x+47y = x^2 - 14x + 47\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)

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Q. Solve the system of equations.\newliney=28x2y = -28x - 2\newliney=x214x+47y = x^2 - 14x + 47\newlineWrite the coordinates in exact form. Simplify all fractions and radicals.\newline(______,______)
  1. Substitute yy: Substitute yy from the first equation into the second equation. Since y=28x2y = -28x - 2, we can replace yy in the second equation with 28x2-28x - 2. The second equation becomes 28x2=x214x+47-28x - 2 = x^2 - 14x + 47.
  2. Solve for x: Now, we need to solve for x. To do this, we will set the equation to zero by moving all terms to one side: x214x+47=28x+2x^2 - 14x + 47 = 28x + 2. This simplifies to x242x+45=0x^2 - 42x + 45 = 0.
  3. Factor the quadratic equation: Next, we factor the quadratic equation x242x+45=0x^2 - 42x + 45 = 0. The factors of 4545 that add up to 4242 are 11 and 4545, but we need them to subtract to 4242, so the factors are 33 and 1515. This gives us (x3)(x15)=0(x - 3)(x - 15) = 0.
  4. Find solutions for xx: Setting each factor equal to zero gives us the solutions for xx. So, x3=0x - 3 = 0 or x15=0x - 15 = 0. This means x=3x = 3 or x=15x = 15.
  5. Substitute xx back: Now we substitute xx back into the first equation y=28x2y = -28x - 2 to find the corresponding yy values. First, for x=3x = 3, we get y=28(3)2y = -28(3) - 2, which simplifies to y=842y = -84 - 2, so y=86y = -86.
  6. Find corresponding y values: Next, for x=15x = 15, we substitute into the first equation y=28x2y = -28x - 2 to get y=28(15)2y = -28(15) - 2, which simplifies to y=4202y = -420 - 2, so y=422y = -422.
  7. Identify solutions: We now have two solutions for the system of equations: (3,86)(3, -86) and (15,422)(15, -422). These are the coordinates of the points where the two equations intersect.

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