Ask questions and get helpful answers.

How do I state if the point given is a solution to the system of equations.

-x^2 + 2y^2 - 2x + 8y + 5 = 0
-x^2 + 26y^2 -2x + 104y + 77 = 0
Point: (-1, -3)
It is not enough to only substitute the point into the equation.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

2 answers

  1. If - x² + 2 y² - 2 x + 8 y + 5 = 0 and - x² + 26 y² - 2 x + 104 y + 77 = 0

    then

    0 = 0

    - x² + 2 y² - 2 x + 8 y + 5 = - x² + 26 y² - 2 x + 104 y + 77

    Add x² to both sides

    2 y² - 2 x + 8 y + 5 = 26 y² - 2 x + 104 y + 77

    Add 2 x to both sides

    2 y² + 8 y + 5 = 26 y² + 104 y + 77

    Subtract 26 y² + 104 y + 77 to both sides

    2 y² + 8 y + 5 - ( 26 y² + 104 y + 77 ) = 0

    2 y² + 8 y + 5 - 26 y² - 104 y - 77 = 0

    - 24 y² - 96 y - 72 = 0

    Divide both sides by - 24

    y² + 4 y + 3 = 0

    The solutions are:

    y = - 3 and y = - 1

    Put y = - 3 in equation - x² + 2 y² - 2 x + 8 y + 5 = 0

    - x² + 2 ∙ ( - 3 )² - 2 x + 8 ∙ ( - 3 ) + 5 = 0

    - x² + 2 ∙ 9 - 2 x - 24 + 5 = 0

    - x² + 18 - 2 x - 24 + 5 = 0

    - x² - 2 x - 1 = 0

    Multiply both sides by - 1

    x² + 2 x + 1 = 0

    ( x + 1 )² = 0

    x + 1 = ± √ 0

    x + 1 = 0

    x = - 1

    Put y = - 3 in equation - x² + 26 y² - 2 x + 104 y + 77 = 0

    - x² + 26 ∙ ( - 3 )² - 2 x + 104 ∙ ( - 3 ) + 77 = 0

    - x² + 26 ∙ 9 - 2 x - 312 + 77 = 0

    - x² + 234 - 2 x - 312 + 77 = 0

    - x² - 2 x - 1 = 0

    x² + 2 x + 1 = 0

    x² + 2 x + 1 = 0

    ( x + 1 )² = 0

    x + 1 = ± √ 0

    x + 1 = 0

    x = - 1

    Again x = - 1

    So point x = - 1 , y = - 3 is the solutions.

    Put y = - 1 in equation - x² + 2 y² - 2 x + 8 y + 5 = 0

    - x² + 2 ∙ ( - 1 )² - 2 x + 8 ∙ ( - 1 ) + 5 = 0

    - x² + 2 ∙ 1 - 2 x - 8 + 5 = 0

    - x² + 2 - 2 x - 8 + 5 = 0

    - x² - 2 x - 1 = 0

    Multiply both sides by - 1

    x² + 2 x + 1 = 0

    ( x + 1 )² = 0

    x + 1 = ± √ 0

    x + 1 = 0

    x = - 1

    Put y = - 1 in equation - x² + 26 y² - 2 x + 104 y + 77 = 0

    - x² + 26 ∙ ( - 1 )² - 2 x + 104 ∙ ( - 1 ) + 77 = 0

    - x² + 26 - 2 x - 104 + 77 = 0

    - x² + 103 - 2 x - 104 = 0

    - x² - 2 x - 1 = 0

    Multiply both sides by - 1

    x² + 2 x + 1 = 0

    ( x + 1 )² = 0

    x + 1 = ± √ 0

    x + 1 = 0

    x = - 1

    Again x = - 1

    So point x = - 1 , y = - 1 is the solutions.

    The solutions are:

    ( - 1 , - 3 ) , ( - 1 , - 1 )

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. first one
    -x^2 + 2y^2 - 2x + 8y + 5 = 0
    - (x^2 + 2 x + 1 ) +1 + 2 (y^2 + 4 y+ 4) - 8 + 5 = 0
    (x+1)^2 - 2(y+2)^2 = -2
    second one
    -x^2 + 26y^2 -2x + 104y + 77 = 0
    - (x^2 +2 x + 1) + 1 + 26 ( y^2 + 4 y + 4) - 104 +77= 26
    (x+1)^2 - 26 (y+2)^2 = -26
    LOL, look when x = -1 and y = -3

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.