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{:[-3y-4=9x+2],[y-4x-2x^(2)=3]:}
If 
(x,y) is a solution to the system of equations shown, which of the following are 
x-coordinates of the solutions?
Choose 1 answer:
(A) 
(5)/(2) and 1
(B) 
-(5)/(2) and -1
(C) -1 and 1
(D) 
-(5)/(2) and 
(11)/(2)

3y4amp;=9x+2y4x2x2amp;=3 \begin{aligned} -3 y-4 & =9 x+2 \\ y-4 x-2 x^{2} & =3 \end{aligned} \newlineIf (x,y) (x, y) is a solution to the system of equations shown, which of the following are x x -coordinates of the solutions?\newlineChoose 11 answer:\newline(A) 52 \frac{5}{2} and 11\newline(B) 52 -\frac{5}{2} and 1-1\newline(C) 1-1 and 11\newline(D) 52 -\frac{5}{2} and 112 \frac{11}{2}

Full solution

Q. 3y4=9x+2y4x2x2=3 \begin{aligned} -3 y-4 & =9 x+2 \\ y-4 x-2 x^{2} & =3 \end{aligned} \newlineIf (x,y) (x, y) is a solution to the system of equations shown, which of the following are x x -coordinates of the solutions?\newlineChoose 11 answer:\newline(A) 52 \frac{5}{2} and 11\newline(B) 52 -\frac{5}{2} and 1-1\newline(C) 1-1 and 11\newline(D) 52 -\frac{5}{2} and 112 \frac{11}{2}
  1. Write Equations: Write down the system of equations.\newline3y4=9x+2-3y - 4 = 9x + 2\newliney4x2x2=3y - 4x - 2x^2 = 3\newlineWe need to solve this system to find the xx-coordinates of the solutions.
  2. Solve for y: Solve the first equation for y.\newline3y=9x+6-3y = 9x + 6\newliney=(9x+6)/3y = -(9x + 6)/3\newliney=3x2y = -3x - 2\newlineNow we have yy in terms of xx.
  3. Substitute yy into 22nd equation: Substitute the expression for yy from Step 22 into the second equation.\newline3x24x2x2=3-3x - 2 - 4x - 2x^2 = 3\newlineCombine like terms.\newline2x27x2=3-2x^2 - 7x - 2 = 3
  4. Set equation to zero: Move all terms to one side to set the equation to zero.\newline2x27x23=0-2x^2 - 7x - 2 - 3 = 0\newline2x27x5=0-2x^2 - 7x - 5 = 0\newlineNow we have a quadratic equation.
  5. Factor or use quadratic formula: Factor the quadratic equation, if possible.\newlineThe quadratic equation 2x27x5=0-2x^2 - 7x - 5 = 0 does not factor easily, so we will use the quadratic formula to find the roots.\newlineThe quadratic formula is x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where aa, bb, and cc are the coefficients from the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.\newlineFor our equation, a=2a = -2, b=7b = -7, and c=5c = -5.
  6. Calculate discriminant: Calculate the discriminant b24acb^2 - 4ac.\newlineDiscriminant D=(7)24(2)(5)D = (-7)^2 - 4(-2)(-5)\newlineD=4940D = 49 - 40\newlineD=9D = 9\newlineSince the discriminant is positive, we will have two real solutions.
  7. Use quadratic formula for xx: Use the quadratic formula to find the xx-coordinates.x=(7)±92(2)x = \frac{-(-7) \pm \sqrt{9}}{2(-2)}x=7±34x = \frac{7 \pm 3}{-4}Now we have two solutions for xx.
  8. Calculate solutions for x: Calculate the two solutions for x.\newlineFirst solution:\newlinex=7+34x = \frac{7 + 3}{-4}\newlinex=104x = \frac{10}{-4}\newlinex=52x = -\frac{5}{2}\newlineSecond solution:\newlinex=734x = \frac{7 - 3}{-4}\newlinex=44x = \frac{4}{-4}\newlinex=1x = -1\newlineThe x-coordinates of the solutions are 52-\frac{5}{2} and 1-1.

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