The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

The original number could be 85.

Step-by-step explanation:

Let the 2 digits be x and y.

Let the number be xy then, assuming that x is the larger digit:

x - y = 3.

x = y + 3

Also

10y + x + 10x + y = 143

Substituting for x:

10y + y + 3 + 10(y + 3) + y = 143

22y + 33 = 143

22y = 110

y = 5.

So x = y + 3 = 8.

10 you would take 50/ .2

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)