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Zorluk: Çok zorSimultaneous Linear and Quadratic Equations

If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the real solution pairs to the simultaneous equations 3x−y=53x - y = 5 and x2+2xy−y2=7x^2 + 2xy - y^2 = 7, with x1<x2x_1 < x_2, what is the value of y2−y1y_2 - y_1?

  1. 18Cevap
  2. B
    6
  3. C
    20
  4. D
    14

Cevap

18
Expressing yy as 3x−53x - 5 and substituting into the quadratic equation yields x2−10x+16=0x^2 - 10x + 16 = 0. Solving gives x1=2x_1 = 2 and x2=8x_2 = 8. Evaluating y=3x−5y = 3x - 5 for both values gives y1=1y_1 = 1 and y2=19y_2 = 19. The difference y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18.

Adım Adım Çözüm

1
Express yy in terms of xx from the linear equation
y=3x−5y = 3x - 5
Isolation of yy facilitates substitution into the non-linear equation.
2
Substitute y=3x−5y = 3x - 5 into the quadratic equation x2+2xy−y2=7x^2 + 2xy - y^2 = 7
x2+2x(3x−5)−(3x−5)2=7x^2 + 2x(3x - 5) - (3x - 5)^2 = 7
Reduces the system to a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression
x2+6x2−10x−(9x2−30x+25)=7  ⟹  −2x2+20x−32=0  ⟹  x2−10x+16=0x^2 + 6x^2 - 10x - (9x^2 - 30x + 25) = 7 \implies -2x^2 + 20x - 32 = 0 \implies x^2 - 10x + 16 = 0
Simplifies the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation x2−10x+16=0x^2 - 10x + 16 = 0 by factorization
(x−2)(x−8)=0  ⟹  x1=2(x - 2)(x - 8) = 0 \implies x_1 = 2 and x2=8x_2 = 8
Identifies the two roots with x1<x2x_1 < x_2 as specified by the condition.
5
Determine the corresponding yy-values using y=3x−5y = 3x - 5
y1=3(2)−5=1y_1 = 3(2) - 5 = 1 and y2=3(8)−5=19y_2 = 3(8) - 5 = 19
Obtains the complete coordinate solution pairs (2,1)(2, 1) and (8,19)(8, 19).
6
Calculate the required difference y2−y1y_2 - y_1
y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18
Computes the final required target value.

Anahtar Kavram

Solving Simultaneous Linear and Quadratic Equations by Substitution
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Simultaneous Linear and Quadratic Equations

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