# Verify if the point (-1,-3) is a solution for the system: -x^2 + 2y^2 - 2x + 8y + 5 = 0 -x^2 + 26y^2 - 2x + 104y + 77 = 0

hala718 | High School Teacher | (Level 1) Educator Emeritus

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-x^2 + 2y^2 - 2x + 8y + 5 = 0.............(1)

-x^2 + 26y^2 - 2x + 104y + 77 = 0...............(1)

To verify if the point (-1, -3) is a solution for the system, then the point should verify both equations:

Let us substritute in equation (1):

-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

25 - 25 = 0

Then the point (-1, -3) is a solution for equation (1):

Now we will substitute in equation (2):

-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

- 313 + 313 = 0

Then the point (-1, -3) is a solution for equation (2):

Then the point (-1,3) is a solution for the system.

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find if (-1, -3) satisfies the system of equations:

-x^2 + 2y^2 - 2x + 8y + 5 = 0

-x^2 + 26y^2 - 2x + 104y + 77 = 0

To do that we just need to substitute x= -1 and y = -3 in the two expressions and see if the result is equal to zero.

Substituting (-1, -3) in the first expression

-x^2 + 2y^2 - 2x + 8y + 5

=> -(-1)^2 + 2(-3)^2 + 2 - 8*3 + 5

=> -1 + 18 + 2 - 24 + 5

= 0

Similarly for the second expression

-x^2 + 26y^2 - 2x + 104y + 77

=> -(-1)^2 + 26*(-3)^2 + 2 - 104*3 + 77

=> -1 + 234 +2 - 312 + 77

=0

Therefore ( -1, -3) satisfies the system of equations.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To verify if the pair (-1,-3) is the solution of the system, we have to substitute x and y by the values -1 and -3 to check if they cancel the equations.

We'll substitute (-1,-3) in the first equation:

-(-1)^2 + 2(-3)^2 - 2(-1) + 8(-3) + 5 = 0

We'll raise to square and remove the brackets:

-1 + 18 + 2 - 24 + 5 = 0

We'll compute and we'll get:

(18 + 2 + 5) - (24 + 1) = 0

25 - 25 = 0

0 = 0

The pair (-1,-3) is the solution of the first equation.

Now, we'll substitute (-1,-3) in the 2nd equation:

-x^2 + 26y^2 - 2x + 104y + 77 = 0 -(-1)^2 + 26(-3)^2 - 2(-1) + 104(-3) + 77 = 0

We'll raise to square and remove the brackets:

-1 + 234 + 2 - 312 + 77 = 0

312 - 312 = 0

0 = 0

The pair (-1,-3) is the solution of the 2nd equation, too.