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y=2-3x y=5x^(2) If (x,y) is a solution to the system of equations shown and x > 0, what is the value of x ?

y=2−3x y=2-3 x \newliney=5x2 y=5 x^{2} \newlineIf (x,y) (x, y) is a solution to the system of equations shown and x>0 , what is the value of x x ?

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Composition of linear and quadratic functions: find a valueright chevron icon

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Q. y=2−3x y=2-3 x \newliney=5x2 y=5 x^{2} \newlineIf (x,y) (x, y) is a solution to the system of equations shown and x>0 x>0 , what is the value of x x ?
  1. Set Equations Equal: We have a system of two equations:\newline11. y=2−3xy = 2 - 3x\newline22. y=5x2y = 5x^2\newlineTo find the value of xx, we need to set the two equations equal to each other because they both equal yy.
  2. Quadratic Equation: Set the two equations equal to each other:\newline2−3x=5x22 - 3x = 5x^2\newlineThis gives us a quadratic equation.
  3. Rearrange to Standard Form: Rearrange the equation to set it to zero:\newline5x2+3x−2=05x^2 + 3x - 2 = 0\newlineNow we have a standard quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  4. Solve Quadratic Equation: To solve the quadratic equation, we can either factor it, complete the square, or use the quadratic formula. The quadratic formula is x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Let's try to factor it first.
  5. Factor or Use Formula: Attempt to factor the quadratic equation:\newline5x2+3x−2=(5x−1)(x+2)5x^2 + 3x - 2 = (5x - 1)(x + 2)\newlineHowever, this factorization is incorrect because it does not yield the original quadratic when multiplied out. Therefore, we should use the quadratic formula instead.

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