A board is 65 inches long into two pieces, so that one piece is 7 inches shorter than twice the length of the other pieces. Find the length of two pieces

24 in, 41 in

Step-by-step explanation:

Set shorter piece is x in, and another longer one is y.

1. x + y = 65

2. y = 2x - 7

plug in the second equation into y of the first equation, we get:

x + (2x - 7) = 65

x + 2x - 7 = 65

    3x - 7 = 65  （Add 7 to the both side)

          3x = 72     (Divide 3 to both side)

            x = 24 in

Plug x in the second equations to slove y.

y = 2x - 7

y = 2(24) - 7

y = 41 in

< 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)