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Zorluk: ZorSimultaneous Linear and Quadratic Equations

Consider the system of simultaneous equations x3y=2x - 3y = 2 and x22xy4y2=19x^2 - 2xy - 4y^2 = 19. What is the sum of the xx-values of the real solution pairs (x,y)(x, y)?

  1. 2828Cevap
  2. B
    88
  3. C
    2020
  4. D
    20-20

Cevap

The sum of the xx-values of the real solution pairs is 2828.
Rearranging the linear equation gives x=3y+2x = 3y + 2. Substituting this into x22xy4y2=19x^2 - 2xy - 4y^2 = 19 results in (3y+2)22(3y+2)y4y2=19(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19, which simplifies to y28y+15=0y^2 - 8y + 15 = 0. The roots are y=5y = 5 and y=3y = 3. Substituting these back into x=3y+2x = 3y + 2 yields x=17x = 17 and x=11x = 11. Their sum is 17+11=2817 + 11 = 28.

Adım Adım Çözüm

1
Express xx in terms of yy using the linear equation.
x=3y+2x = 3y + 2
Isolation of xx allows direct substitution into the quadratic equation.
2
Substitute x=3y+2x = 3y + 2 into the quadratic equation x22xy4y2=19x^2 - 2xy - 4y^2 = 19.
(3y+2)22(3y+2)y4y2=19(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19
This reduces the non-linear system to a single quadratic equation in terms of yy.
3
Expand and simplify the quadratic equation.
(9y2+12y+4)(6y2+4y)4y2=19    y2+8y+4=19    y28y+15=0(9y^2 + 12y + 4) - (6y^2 + 4y) - 4y^2 = 19 \implies -y^2 + 8y + 4 = 19 \implies y^2 - 8y + 15 = 0
Putting the quadratic expression into standard form ay2+by+c=0ay^2 + by + c = 0 facilitates finding its roots.
4
Solve the quadratic equation y28y+15=0y^2 - 8y + 15 = 0 for yy.
(y5)(y3)=0    y1=5,y2=3(y - 5)(y - 3) = 0 \implies y_1 = 5, y_2 = 3
Factoring determines the ordinate values for the solution pairs.
5
Calculate the corresponding xx-values and find their sum.
For y1=5y_1 = 5: x1=3(5)+2=17x_1 = 3(5) + 2 = 17.
For y2=3y_2 = 3: x2=3(3)+2=11x_2 = 3(3) + 2 = 11.
Sum = 17+11=2817 + 11 = 28.
Plugging the yy-values back into x=3y+2x = 3y + 2 gives the abscissas, which are added to answer the question.

Anahtar Kavram

Solving simultaneous linear and quadratic equations by algebraic substitution
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