# 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.

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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 )

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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

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