Which statements are true about the ordered pair (5
- 6) and the system of equations?
x + y = -1
| 3x - y = 21
select each correct answer.
- 6) is a solution to the first equation because it makes the
the ordered pair (5
first equation true.
the ordered pair (5 - 6) is a solution to the second equation because it makes
the second equation true.
the ordered pair (5 - 6) is not a solution to the system because it makes at least
one of the equations false.
the ordered pair (5,
equations true.
- 6) is a solution to the system because it makes both

The ordered pair (5,-6) is a solution to the first equation because it makes the first equation true

The ordered pair (5 - 6) is a solution to the second equation because it makes  the second equation true

The ordered pair (5 - 6) is a solution to the system because it makes both  equations true

Step-by-step explanation:

we have

--- equation A

----> equation B

we know that

If a ordered pair is a solution of the system of equations, then the ordered pair must satisfy both equations

Solve the system of equations by elimination

Adds equation A and equation B

Find the value of y

substitute the value of x in any equation

The solution of the system is the point (5,-6)

therefore

Verify each statement

a) The ordered pair (5,-6) is a solution to the first equation because it makes the first equation true.

The statement is true (see the procedure)

b) The ordered pair (5 - 6) is a solution to the second equation because it makes  the second equation true

The statement is true (see the procedure)

c) The ordered pair (5 - 6) is not a solution to the system because it makes at least  one of the equations false

The statement is false (see the procedure)

d) The ordered pair (5 - 6) is a solution to the system because it makes both  equations true

The statement is true (see the procedure)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)