The sum of ages of Noi's and Noy's is 26 years. The different between four times Noi's age and two times Noy's age is 28 years. Find the age of Noi and Noy. Write as an equation

Let the age will be x

4x+2x-28 years

6x-28

x-28 by 6 -4.66

SO THE ANSWER IS 4.66

Noi = x

Noy = y

x+y = 26 ... (1)

4x-2y= 28 ... (2)

Solve:

x+y= 26

x= 26-y (keep it)

4x-2y= 28

4(26-y)-2y=28

104-4y-2y= 28

-6y= 28-104

-6y=-76

y= 12,6 (12 years 6 mouths)

x+y= 26

x + 12.6 = 26

x= 26-12.6

x= 13, 4 ( 13 years 4 mouths)

< 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)