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Difficulty: MediumNonlinear Systems of Equations
y=x2−5x+8y = x^2 - 5x + 8
y=2x+2y = 2x + 2

The system of equations above has two real solutions, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). If y1>y2y_1 > y_2, what is the value of x1−x2x_1 - x_2?

  1. A
    -5
  2. 5Answer
  3. C
    7
  4. D
    10

Answer

The value of the difference between the two x-coordinates is 5.
To solve the system, substitute y=2x+2y = 2x + 2 into y=x2−5x+8y = x^2 - 5x + 8 to obtain 2x+2=x2−5x+82x + 2 = x^2 - 5x + 8. Subtracting 2x+22x + 2 from both sides gives the quadratic equation x2−7x+6=0x^2 - 7x + 6 = 0. Factoring this equation yields (x−6)(x−1)=0(x - 6)(x - 1) = 0, which gives the x-coordinates x=6x = 6 and x=1x = 1. Substituting these back into y=2x+2y = 2x + 2 gives the corresponding y-coordinates: y=14y = 14 when x=6x = 6, and y=4y = 4 when x=1x = 1. Thus, the two solutions are (6,14)(6, 14) and (1,4)(1, 4). Because y1>y2y_1 > y_2, we must have (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and (x2,y2)=(1,4)(x_2, y_2) = (1, 4). The value of x1−x2x_1 - x_2 is therefore 6−1=56 - 1 = 5.

Step-by-Step Solution

1
Substitute the expression for yy from the second equation into the first equation.
2x+2=x2−5x+82x + 2 = x^2 - 5x + 8
This substitution reduces the system of two equations to a single quadratic equation in terms of xx.
2
Rearrange the quadratic equation into standard form by subtracting 2x2x and 22 from both sides.
x2−7x+6=0x^2 - 7x + 6 = 0
Putting the equation in standard form is necessary to factor it and find its roots.
3
Factor the quadratic equation to find the two possible values of xx.
(x−6)(x−1)=0(x - 6)(x - 1) = 0, which gives x=6x = 6 or x=1x = 1.
Factoring allows us to find the x-coordinates of the points where the two graphs intersect.
4
Substitute the x-values back into the linear equation y=2x+2y = 2x + 2 to find their corresponding y-values.
For x=6x = 6, y=2(6)+2=14y = 2(6) + 2 = 14. For x=1x = 1, y=2(1)+2=4y = 2(1) + 2 = 4. The two solutions are (6,14)(6, 14) and (1,4)(1, 4).
Finding the y-values helps identify which coordinate pair corresponds to (x1,y1)(x_1, y_1) and which to (x2,y2)(x_2, y_2) using the given condition.
5
Apply the condition y1>y2y_1 > y_2 to assign the variables and calculate x1−x2x_1 - x_2.
Since 14>414 > 4, the solution with the larger y-value is (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and the other is (x2,y2)=(1,4)(x_2, y_2) = (1, 4). Thus, x1−x2=6−1=5x_1 - x_2 = 6 - 1 = 5.
This calculation yields the final requested value.

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.
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Topics

Nonlinear Systems of Equations

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