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Difficulty: EasySystems of Linear and Non-Linear Equations

A system of equations is given below:

y=(x−1)2−6y=x−1\begin{aligned} y &= (x - 1)^2 - 6 \\ y &= x - 1 \end{aligned}

If (x,y)(x, y) is a solution to this system in the first quadrant, what is the value of x+yx + y?

  1. A
    5
  2. B
    -3
  3. 7Answer
  4. D
    4
  5. E
    9

Answer

The sum of the coordinates of the first quadrant solution is 7.
To solve the system, substitute the expression for yy from the linear equation into the quadratic equation to get x−1=(x−1)2−6x - 1 = (x - 1)^2 - 6. Substituting u=x−1u = x - 1 yields u2−u−6=0u^2 - u - 6 = 0. Factoring the quadratic expression gives (u−3)(u+2)=0(u - 3)(u + 2) = 0, so u=3u = 3 or u=−2u = -2. Since u=x−1u = x - 1, this means x=4x = 4 or x=−1x = -1. Evaluating both cases gives the intersection points (4,3)(4, 3) and (−1,−2)(-1, -2). Only (4,3)(4, 3) is in the first quadrant. Adding these coordinates together yields 4+3=74 + 3 = 7.

Step-by-Step Solution

1
Substitute y=x−1y = x - 1 into the quadratic equation to set up an equation in terms of xx.
x−1=(x−1)2−6x - 1 = (x - 1)^2 - 6
To find the xx-coordinates of the intersection points.
2
Solve the equation for xx by substituting u=x−1u = x - 1.
u=u2−6⇒u2−u−6=0⇒(u−3)(u+2)=0u = u^2 - 6 \Rightarrow u^2 - u - 6 = 0 \Rightarrow (u - 3)(u + 2) = 0. This gives u=3u = 3 or u=−2u = -2.
To find the values of the substituted variable uu.
3
Find the corresponding xx and yy values for both cases.
Case 1: x−1=3⇒x=4x - 1 = 3 \Rightarrow x = 4, which gives y=3y = 3. Point is (4,3)(4, 3). Case 2: x−1=−2⇒x=−1x - 1 = -2 \Rightarrow x = -1, which gives y=−2y = -2. Point is (−1,−2)(-1, -2).
To determine the full coordinates of all intersection points.
4
Identify the first quadrant solution and calculate the sum of its coordinates.
The first quadrant solution is (4,3)(4, 3). The sum of the coordinates is 4+3=74 + 3 = 7.
To answer the question's requirement for the sum of coordinates in the first quadrant.

Key Concept

Solving systems of linear and quadratic equations by substitution and identifying quadrant-specific solutions.
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Topics

Systems of Linear and Non-Linear Equations

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