A and B are friends and …

Let the ages of A and B be x and y years respectively.
Then, age difference of A and B is 2 years.
{tex}\Rightarrow{/tex}x - y = {tex}\pm{/tex}2
A's father D is twice as old as A and B is twice as old as his sister C.
Then, D's age =2x years. and, C's age {tex}= \frac { y } { 2 }{/tex}years.
Clearly, D is older than C
The age of D and C differ by 40 years.
{tex}\therefore{/tex} {tex}2 x - \frac { y } { 2 } = 40 {/tex}
{tex}\Rightarrow 4 x - y = 80{/tex}
Thus, we have the following two systems of linear equations
x - y =2 ...............(i)
and, 4x - y =80 ...........(ii)
x - y = -2 ............(iii)
and, 4x - y=80 ..................(iv)
Subtracting equation (i) from equation (ii), we get
(x - y) - (4x - y)= 2 - 80
⇒ x - y - 4x + y = -78
⇒ - 3x = - 78
⇒ 3x =78 
⇒x =26
Putting x =26 in equation (i), we get
x - y =2
⇒ 26 - y = 2
⇒ y =24
Subtracting equation (iv) from equation (iii), we get
{tex}- 3 x = - 82 {/tex}
{tex}\Rightarrow x = \frac { 82 } { 3 } = 27 \frac { 1 } { 3 }{/tex}
Putting  {tex}x = \frac { 82 } { 3 }{/tex}in equation (iii), we get
{tex}y = \frac { 82 } { 3 } + 2 = \frac { 88 } { 3 } = 29 \frac { 1 } { 3 }{/tex}
Hence, A's age =26 years and B's age =24 years
or,
A's age {tex}= 27 \frac { 1 } { 3 }{/tex}Years and B's age {tex}29 \frac { 1 } { 3 }{/tex}years.