4x-3y=7 x=13-3y Solve for the variables.

x = 4

y = 3

Step-by-step explanation:

4x-3y=7

x=13-3y

Put x as 13-3y in the first equation.

4(13-3y) - 3y= 7

52 -12y - 3y = 7

-15y = 7 - 52

-15y = -45

y = 3

Put y as 3 in the second equation.

x = 13-3(3)

x = 13-9

x = 4

x = 4, y = 3

Step-by-step explanation:

4x-3y = 7

Putting x = 13-3y in the above equation

=> 4(13-3y)-3y = 7

=> 52-12y-3y = 7

=> 52-7 = 15y

=> 45 = 15y

=> y = 3

Now, x

=> x = 13-3y

=> x = 13-3(3)

=> x = 13-9

=> x = 4

what shape is

< 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)