Solve the system using elimination. 3 x plus 3 y equals 27. x minus 3 y equals negative 11.

- 5/4

Step-by-step explanation:

x=4 and y=5 commonly notated as (4,5)

Step-by-step explanation:

The equations

Equation 1: 3x+3y=27

Equation 2: x-3y= -11

The second equation looks like it will be the easiest to rearrange.

x-3y = -11

add 3y to each side to get

x = 3y-11

Substitute that in for x in equation 1 so

3x+3y=27

turns into

3(3y-11) +3y =27

multiply out to get

9y-33+3y =27

Condense variables

12y-33=27

add 33 to each side

12y= 60

Divide by 12

y = 5

Plug into one of the original equations (I will be doing both to show it doesn't matter what equation you use, but you only need to use one equation, I personally think it is easier to plug into the second equation)

Equation 1: 3x+3y=27

Plug in what we got for y

3x+3(5)=27

multiply

3x+15=27

subtract 15 from each side

3x=12

divide by 3

x=4

Equation 2: x-3y = -11

Plug in what we got for y

x-3(5) =-11

Multiply

x-15= -11

Add 15 to each side

x=4

the answer would be c:   ax + b = ax + c

step-by-step explanation:

< 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)