Question

Difficulty: MediumSimultaneous Linear and Quadratic Equations

What are the values of yy for the real solution pairs (x,y)(x, y) that satisfy the simultaneous equations y3x=2y - 3x = 2 and y=x2x+5y = x^2 - x + 5?

  1. 55 or 1111Answer
  2. B
    11 or 33
  3. C
    1-1 or 7-7
  4. D
    33 or 55

Answer

55 or 1111
Substituting y=3x+2y = 3x + 2 into the quadratic equation y=x2x+5y = x^2 - x + 5 gives 3x+2=x2x+53x + 2 = x^2 - x + 5, which simplifies to x24x+3=0x^2 - 4x + 3 = 0. Solving this yields x=1x = 1 and x=3x = 3. Substituting these into the linear equation gives y=3(1)+2=5y = 3(1) + 2 = 5 and y=3(3)+2=11y = 3(3) + 2 = 11. Therefore, the possible values of yy are 55 or 1111.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation
y=3x+2y = 3x + 2
Linear equations can easily be substituted into quadratic equations.
2
Equate the linear expression for yy to the quadratic equation
3x+2=x2x+53x + 2 = x^2 - x + 5
Both expressions represent yy.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x24x+3=0x^2 - 4x + 3 = 0
Subtract 3x3x and 22 from both sides.
4
Factorize and solve for xx
(x1)(x3)=0    x=1 or x=3(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3
Finding the roots of the quadratic equation gives the xx-coordinates of the solution pairs.
5
Substitute each xx-value back into y=3x+2y = 3x + 2 to find yy
When x=1x = 1, y=3(1)+2=5y = 3(1) + 2 = 5. When x=3x = 3, y=3(3)+2=11y = 3(3) + 2 = 11.
The question specifically asks for the values of yy.

Key Concept

Simultaneous Linear and Quadratic Equations
Estimated Time:1m 30s
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