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Zorluk: 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 x−2y=1x - 2y = 1 and x2−xy+y2=7x^2 - xy + y^2 = 7, what is the value of x1x2+y1y2x_1 x_2 + y_1 y_2?

  1. A
    −9-9
  2. −11-11Cevap
  3. C
    −7-7
  4. D
    −1-1

Cevap

The value of x1x2+y1y2x_1 x_2 + y_1 y_2 is −11-11.
Substituting x=2y+1x = 2y + 1 into x2−xy+y2=7x^2 - xy + y^2 = 7 gives 3y2+3y−6=03y^2 + 3y - 6 = 0, which reduces to y2+y−2=0y^2 + y - 2 = 0. The roots are y=1y = 1 and y=−2y = -2. Substituting back into the linear expression yields corresponding values x=3x = 3 and x=−3x = -3. The two solution pairs are (3,1)(3, 1) and (−3,−2)(-3, -2). Calculating x1x2+y1y2=(3)(−3)+(1)(−2)x_1 x_2 + y_1 y_2 = (3)(-3) + (1)(-2) yields −11-11.

Adım Adım Çözüm

1
Express xx in terms of yy from the linear equation.
x=2y+1x = 2y + 1
Isolating variable xx allows direct substitution into the quadratic equation.
2
Substitute x=2y+1x = 2y + 1 into the quadratic equation x2−xy+y2=7x^2 - xy + y^2 = 7.
(2y+1)2−(2y+1)y+y2=7  ⟹  3y2+3y−6=0(2y + 1)^2 - (2y + 1)y + y^2 = 7 \implies 3y^2 + 3y - 6 = 0
This simplifies the system to a single quadratic equation in yy.
3
Solve the quadratic equation 3y2+3y−6=03y^2 + 3y - 6 = 0.
y2+y−2=0  ⟹  (y+2)(y−1)=0y^2 + y - 2 = 0 \implies (y + 2)(y - 1) = 0, giving y1=1y_1 = 1 and y2=−2y_2 = -2
Factoring determines the two possible yy-coordinates.
4
Determine corresponding xx-coordinates for each yy-value.
For y1=1y_1 = 1, x1=2(1)+1=3  ⟹  (3,1)x_1 = 2(1) + 1 = 3 \implies (3, 1). For y2=−2y_2 = -2, x2=2(−2)+1=−3  ⟹  (−3,−2)x_2 = 2(-2) + 1 = -3 \implies (-3, -2).
Substituting each yy into x=2y+1x = 2y + 1 yields the complete solution pairs.
5
Compute x1x2+y1y2x_1 x_2 + y_1 y_2.
(3)(−3)+(1)(−2)=−9−2=−11(3)(-3) + (1)(-2) = -9 - 2 = -11
Evaluates the targeted algebraic expression.

Anahtar Kavram

Solving simultaneous linear and quadratic equations using algebraic substitution.
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