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Zorluk: OrtaSystems of Linear and Non-Linear Equations

A system of equations consists of the linear equation y=x−2y = x - 2 and the quadratic equation y=(x−3)2−5y = (x - 3)^2 - 5. If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the two distinct real solutions to this system, what is the value of the product y1y2y_1 y_2?

  1. A
    -10
  2. -4Cevap
  3. C
    3
  4. D
    4
  5. E
    6

Cevap

The product of the two yy-coordinates of the intersection points is −4-4.
Substituting the linear expression for yy into the quadratic equation yields the quadratic equation x2−7x+6=0x^2 - 7x + 6 = 0. Factoring this equation gives the xx-coordinates 66 and 11. Substituting these values back into the linear equation y=x−2y = x - 2 yields the corresponding yy-coordinates 44 and −1-1. The product of these yy-coordinates is 4×(−1)=−44 \times (-1) = -4.

Adım Adım Çözüm

1
Equate the two expressions for yy to solve for the xx-coordinates of the intersection points.
x−2=(x−3)2−5x - 2 = (x - 3)^2 - 5
At the points of intersection, the yy-values of both equations are equal.
2
Expand the squared binomial and simplify the equation into standard quadratic form.
x2−7x+6=0x^2 - 7x + 6 = 0
Expanding (x−3)2(x - 3)^2 yields x2−6x+9x^2 - 6x + 9. Substituting this back gives x−2=x2−6x+4x - 2 = x^2 - 6x + 4. Subtracting xx and adding 22 to both sides yields the standard quadratic equation.
3
Factor the quadratic equation to find the solutions for xx.
x=6x = 6 or x=1x = 1
The equation factors as (x−6)(x−1)=0(x - 6)(x - 1) = 0, giving the two xx-values.
4
Substitute each xx-value back into the linear equation y=x−2y = x - 2 to find the corresponding yy-coordinates.
y1=4y_1 = 4 and y2=−1y_2 = -1
For x=6x = 6, y=6−2=4y = 6 - 2 = 4. For x=1x = 1, y=1−2=−1y = 1 - 2 = -1.
5
Calculate the product of the two yy-coordinates.
y1y2=4×(−1)=−4y_1 y_2 = 4 \times (-1) = -4
The question asks for the product of the two yy-coordinates.

Anahtar Kavram

Solving a system of linear and quadratic equations using substitution and factoring.

Alternatif Yöntem

Alternatively, you can expand the quadratic equation first to y=x2−6x+4y = x^2 - 6x + 4 and set it equal to the linear equation y=x−2y = x - 2. Solving for xx yields x2−7x+6=0x^2 - 7x + 6 = 0, from which you can find the coordinates and calculate their product.
Tahmini Süre:1m 30s
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Konular

Systems of Linear and Non-Linear Equations

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