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Difficulty: MediumNonlinear Systems of Equations
y=2x2−5x−1y=3x−7\begin{aligned} y &= 2x^2 - 5x - 1 \\ y &= 3x - 7 \end{aligned}

If (x,y)(x, y) is a solution to the system of equations above and y>0y > 0, what is the value of x+yx + y?

  1. A
    -11
  2. 5Answer
  3. C
    -3
  4. D
    1

Answer

5
Substituting y=3x−7y = 3x - 7 into y=2x2−5x−1y = 2x^2 - 5x - 1 yields the quadratic equation 3x−7=2x2−5x−13x - 7 = 2x^2 - 5x - 1. Moving all terms to one side gives 2x2−8x+6=02x^2 - 8x + 6 = 0. Dividing by 2 simplifies this to x2−4x+3=0x^2 - 4x + 3 = 0, which factors as (x−1)(x−3)=0(x - 1)(x - 3) = 0. This gives the solutions x=1x = 1 and x=3x = 3. Substituting these into the linear equation yields the coordinate points (1,−4)(1, -4) and (3,2)(3, 2). Since the problem states that y>0y > 0, we choose the solution (3,2)(3, 2). The value of x+yx + y is 3+2=53 + 2 = 5.

Step-by-Step Solution

1
Substitute the expression for yy from the linear equation into the quadratic equation.
3x−7=2x2−5x−13x - 7 = 2x^2 - 5x - 1
Since both equations are solved for yy, they can be set equal to each other to find the xx-coordinates of the intersection points.
2
Rearrange the terms to set the quadratic equation equal to zero, and then divide by the common factor.
2x2−8x+6=02x^2 - 8x + 6 = 0, which simplifies to x2−4x+3=0x^2 - 4x + 3 = 0
Putting the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0 allows us to factor and solve for xx.
3
Factor the quadratic equation to find the values of xx.
(x−1)(x−3)=0(x - 1)(x - 3) = 0, so x=1x = 1 or x=3x = 3
Factoring allows us to find the roots of the equation directly.
4
Substitute the xx values back into the linear equation to solve for the corresponding yy values.
For x=1x = 1, y=3(1)−7=−4y = 3(1) - 7 = -4, giving the solution (1,−4)(1, -4). For x=3x = 3, y=3(3)−7=2y = 3(3) - 7 = 2, giving the solution (3,2)(3, 2).
Finding the yy-coordinates completes the solutions (x,y)(x, y) to the system.
5
Apply the given constraint y>0y > 0 and calculate x+yx + y.
The solution (3,2)(3, 2) satisfies y>0y > 0 since 2>02 > 0. Thus, x+y=3+2=5x + y = 3 + 2 = 5.
The problem asks for the sum of the coordinates of the solution that has a positive yy-value.

Key Concept

Solving a system consisting of a linear equation and a quadratic equation by substitution and applying coordinate constraints.
Estimated Time:1m 30s
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Topics

Nonlinear Systems of Equations

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