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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2 answers

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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answered by Bosnian 
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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answered by Anonymous
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