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Difficulty: Very hardSimultaneous Linear and Quadratic Equations

Given the simultaneous equations 2x−y=12x - y = 1 and 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17, let (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) be the real solution pairs such that x1>x2x_1 > x_2. What is the value of 2x1+y22x_1 + y_2?

  1. 1Answer
  2. B
    3
  3. C
    9
  4. D
    -3

Answer

1
Substituting y=2x−1y = 2x - 1 into 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17 simplifies correctly to 4x2−4x−15=04x^2 - 4x - 15 = 0. Factorizing gives roots x=52x = \frac{5}{2} and x=−32x = -\frac{3}{2}. Given x1>x2x_1 > x_2, we set x1=52x_1 = \frac{5}{2} and x2=−32x_2 = -\frac{3}{2}. Substituting x2x_2 back into the linear equation gives y2=−4y_2 = -4. Calculating 2x1+y2=2(52)+(−4)=5−4=12x_1 + y_2 = 2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation.
y=2x−1y = 2x - 1
Substitution method requires expressing one variable in terms of the other.
2
Substitute y=2x−1y = 2x - 1 into the quadratic equation 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17.
2x2−x(2x−1)+(2x−1)2+x−(2x−1)=172x^2 - x(2x - 1) + (2x - 1)^2 + x - (2x - 1) = 17
This converts the system into a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression.
2x2−2x2+x+4x2−4x+1+x−2x+1=17  ⟹  4x2−4x+2=17  ⟹  4x2−4x−15=02x^2 - 2x^2 + x + 4x^2 - 4x + 1 + x - 2x + 1 = 17 \implies 4x^2 - 4x + 2 = 17 \implies 4x^2 - 4x - 15 = 0
Combining like terms reveals the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation 4x2−4x−15=04x^2 - 4x - 15 = 0 by factorization.
(2x−5)(2x+3)=0  ⟹  x=52(2x - 5)(2x + 3) = 0 \implies x = \frac{5}{2} or x=−32x = -\frac{3}{2}
Finding the two real values for xx.
5
Identify x1x_1 and x2x_2 according to x1>x2x_1 > x_2 and compute corresponding yy-values.
x1=52  ⟹  y1=2(52)−1=4x_1 = \frac{5}{2} \implies y_1 = 2\left(\frac{5}{2}\right) - 1 = 4; x2=−32  ⟹  y2=2(−32)−1=−4x_2 = -\frac{3}{2} \implies y_2 = 2\left(-\frac{3}{2}\right) - 1 = -4
Determining the complete coordinate solution pairs.
6
Evaluate the expression 2x1+y22x_1 + y_2.
2(52)+(−4)=5−4=12\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1
Answering the specific value requested in the problem statement.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Estimated Time:2m 30s
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Topics

Simultaneous Linear and Quadratic Equations

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