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{:[y=3x-15],[y=x^(2)-2x-15]:} If (x,y) is a solution to the system of equations and x > 0, what is the value of x ? Choose 1 answer: (A) 1 (B) 5 (c) 6 (D) 15

y=3x−15 y=3 x-15 \newliney=x2−2x−15 y=x^{2}-2 x-15 \newlineIf (x,y) (x, y) is a solution to the system of equations and x>0 , what is the value of x x ?\newlineChoose 11 answer:\newline(A) 11\newline(B) 55\newline(C) 66\newline(D) 1515

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Full solution

Q. y=3x−15 y=3 x-15 \newliney=x2−2x−15 y=x^{2}-2 x-15 \newlineIf (x,y) (x, y) is a solution to the system of equations and x>0 x>0 , what is the value of x x ?\newlineChoose 11 answer:\newline(A) 11\newline(B) 55\newline(C) 66\newline(D) 1515
  1. Set Equations Equal: Set the two equations equal to each other since they both equal yy.3x−15=x2−2x−153x - 15 = x^2 - 2x - 15
  2. Rearrange and Solve: Rearrange the equation to set it to zero and solve for xx.x2−2x−15−3x+15=0x^2 - 2x - 15 - 3x + 15 = 0x2−5x=0x^2 - 5x = 0
  3. Factor Out xx: Factor out xx from the equation.x(x−5)=0x(x - 5) = 0
  4. Solve for x: Solve for x by setting each factor equal to zero.\newlinex=0x = 0 or x−5=0x - 5 = 0
  5. Solve Second Equation: Solve the second equation for xx.x−5=0x - 5 = 0x=5x = 5
  6. Discard x=0x = 0: Since we are given that x > 0, we discard the solution x=0x = 0 and only consider x=5x = 5.

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