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Zorluk: ZorNonlinear Systems of Equations

For a real solution (x,y)(x, y) to the system of equations below, which of the following is a possible value of x+yx + y?

x2+3y2=21x2−2y=5\begin{aligned} x^2 + 3y^2 &= 21 \\ x^2 - 2y &= 5 \end{aligned}
  1. A
    -5
  2. B
    -3
  3. C
    1
  4. 5Cevap

Cevap

The correct answer is 5, which is the sum of the coordinates of the real solution (3, 2) of the system.
Substituting x2=2y+5x^2 = 2y + 5 into the first equation yields the quadratic equation 3y2+2y−16=03y^2 + 2y - 16 = 0. Solving this quadratic equation gives y=2y = 2 or y=−8/3y = -8/3. Since x2x^2 must be non-negative for real solutions, y=−8/3y = -8/3 yields no real solutions because x2=2(−8/3)+5=−1/3<0x^2 = 2(-8/3) + 5 = -1/3 < 0. For y=2y = 2, we have x2=2(2)+5=9x^2 = 2(2) + 5 = 9, which gives x=3x = 3 or x=−3x = -3. The real solutions to the system are (3,2)(3, 2) and (−3,2)(-3, 2). The possible values for x+yx + y are 3+2=53 + 2 = 5 and −3+2=−1-3 + 2 = -1. Therefore, 5 is a possible value of x+yx + y.

Adım Adım Çözüm

1
Rearrange the second equation to express x2x^2 in terms of yy.
x2=2y+5x^2 = 2y + 5
This allows for direct substitution into the first equation to eliminate the xx variable.
2
Substitute the expression for x2x^2 into the first equation and simplify the resulting quadratic equation in terms of yy.
3y2+2y−16=03y^2 + 2y - 16 = 0
Substituting 2y+52y + 5 for x2x^2 in x2+3y2=21x^2 + 3y^2 = 21 gives (2y+5)+3y2=21(2y + 5) + 3y^2 = 21, which simplifies to 3y2+2y−16=03y^2 + 2y - 16 = 0.
3
Factor or use the quadratic formula to solve for yy.
y=2y = 2 or y=−8/3y = -8/3
The quadratic equation factors as (3y+8)(y−2)=0(3y + 8)(y - 2) = 0, giving the roots y=2y = 2 and y=−8/3y = -8/3.
4
Find the corresponding values of xx for each yy and determine the real solutions.
For y=2y = 2, x2=2(2)+5=9  ⟹  x=±3x^2 = 2(2) + 5 = 9 \implies x = \pm 3. For y=−8/3y = -8/3, x2=2(−8/3)+5=−1/3x^2 = 2(-8/3) + 5 = -1/3, which has no real solutions.
A real number squared cannot be negative, so only y=2y = 2 yields real solutions: (3,2)(3, 2) and (−3,2)(-3, 2).
5
Calculate the possible values of x+yx + y for the real solutions.
For (3,2)(3, 2), x+y=3+2=5x + y = 3 + 2 = 5. For (−3,2)(-3, 2), x+y=−3+2=−1x + y = -3 + 2 = -1.
Evaluating the target expression for both valid real solutions gives 5 and -1. Since 5 is the only value present in the options, it is the correct answer.

Anahtar Kavram

Solving systems of nonlinear equations algebraically using substitution and analyzing real solutions.
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Konular

Nonlinear Systems of Equations

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