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Zorluk: OrtaSimultaneous Linear and Quadratic Equations

If 2x+y=72x + y = 7 and x2+xy=6x^2 + xy = 6, what are the possible values of xx?

  1. 11 or 66Cevap
  2. B
    1-1 or 6-6
  3. C
    22 or 33
  4. D
    2-2 or 3-3

Cevap

The possible values of xx are 11 or 66.
From the linear equation 2x+y=72x + y = 7, we get y=72xy = 7 - 2x. Substituting this into x2+xy=6x^2 + xy = 6 gives x2+x(72x)=6x^2 + x(7 - 2x) = 6, which simplifies to x2+7x=6-x^2 + 7x = 6, or x27x+6=0x^2 - 7x + 6 = 0. Factoring gives (x1)(x6)=0(x - 1)(x - 6) = 0, leading to x=1x = 1 or x=6x = 6.

Adım Adım Çözüm

1
Express yy in terms of xx using the linear equation.
y=72xy = 7 - 2x
Isolation of one variable allows substitution into the non-linear equation.
2
Substitute y=72xy = 7 - 2x into the second equation x2+xy=6x^2 + xy = 6.
x2+x(72x)=6    x2+7x2x2=6    x2+7x6=0x^2 + x(7 - 2x) = 6 \implies x^2 + 7x - 2x^2 = 6 \implies -x^2 + 7x - 6 = 0
This reduces the system to a single quadratic equation in xx.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x27x+6=0x^2 - 7x + 6 = 0
Standard quadratic form enables easy factorization.
4
Factor the quadratic equation and solve for xx.
(x1)(x6)=0    x=1 or x=6(x - 1)(x - 6) = 0 \implies x = 1 \text{ or } x = 6
Setting each factor to zero yields the values of xx.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by substitution
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