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.
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 )
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
To determine if a point is a solution to a system of equations, you need to substitute the given point into both equations and see if it satisfies both equations simultaneously. Here are the steps to check if the point (-1, -3) is a solution to the system of equations:
Step 1: Substitute the values of x and y from the given point into the first equation and simplify:
-x^2 + 2y^2 - 2x + 8y + 5 = 0
-(-1)^2 + 2(-3)^2 - 2(-1) + 8(-3) + 5 = 0
-1 + 18 + 2 - 24 + 5 = 0
0 = 0
Step 2: Substitute the values of x and y from the given point into the second equation and simplify:
-x^2 + 26y^2 - 2x + 104y + 77 = 0
-(-1)^2 + 26(-3)^2 - 2(-1) + 104(-3) + 77 = 0
-1 + 234 - 2 - 312 + 77 = 0
-4 = 0
Step 3: Check if the result from both equations is satisfied or not:
Since the result of both equations is not equal to 0 (0 ≠ 0), we can conclude that the point (-1, -3) is not a solution to the system of equations.
It is important to substitute the values into both equations to ensure that the point satisfies all the equations in the system.
To determine if a given point is a solution to a system of equations, you need to substitute the point's coordinates into each equation of the system and check if the equations are satisfied.
Here's how you can check if the point (-1, -3) is a solution to the system of equations:
Step 1: Substitute the coordinates of the point (-1, -3) into the first equation:
-x^2 + 2y^2 - 2x + 8y + 5 = 0
-(-1)^2 + 2(-3)^2 - 2(-1) + 8(-3) + 5 = 0
1 + 18 + 2 - 24 + 5 = 0
2 = 0 (not satisfied)
Step 2: Now, substitute the coordinates of the point (-1, -3) into the second equation:
-x^2 + 26y^2 - 2x + 104y + 77 = 0
-(-1)^2 + 26(-3)^2 - 2(-1) + 104(-3) + 77 = 0
1 + 234 + 2 - 312 + 77 = 0
2 = 0 (not satisfied)
Since both equations don't equal to 0 when using the values (-1, -3), the point (-1, -3) is not a solution to the system of equations.
It is essential to plug in the coordinates of the point into each equation to check if they satisfy the equations individually. Only if all the equations are satisfied, then the point can be considered a solution to the system.