An algebra problem by Panshul Rastogi

$Given\quad that\\ x+y=16\quad ........\quad (i)\\ x-y=10\quad .........\quad (ii)\\ Adding\quad (i)\quad \& \quad (ii),\quad we\quad have\\ 2x=26\quad or\quad \boxed { x=13 } \\ Putting\quad \boxed { x=13 } \quad in\quad (i),\quad we\quad have\\ \boxed { y=3 } \\ \qquad xy=(13)(3)=39\\ Now\quad find\quad { x }^{ 2 }+{ y }^{ 2 }\\ { x }^{ 2 }+{ y }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy-2xy\\ \quad \quad \quad \quad ={ (x+y })^{ 2 }-2xy\\ \quad \quad \quad \quad ={ (16) }^{ 2 }-2(39)\\ \quad \quad \quad \quad =256-78\\ \boxed { { x }^{ 2 }+{ y }^{ 2 }=178 }$

Add the given equations to get:- $2x = 26\\ x = 13$
So, $13 + y = 16\\ y = 3$
So, $x^2 + y^2 = {13}^2 + 3^2 = 169 + 9 = \color{#69047E}{\boxed{178}}$ .

adding the two equations we have $2x=26$ and $x=13$ while subtracting the second from the first we have $2y = 6$ or $y=3$ . $x^2+y^2 = 169 + 9 = \boxed{178}$ .

((X+Y)-(X-Y))/2 or (16-10)/2=3, so y=3, therefore, x=13 and x²+y²=13²+3²=169+9=178

(x+y)^2=16^2=256 x^2+2xy+y^2=256 (x-y)^2=10^2=100 x^2-2xy+y^2=100 (x+y)^2+(x-y)^2=2x^2+2y^2=356 x^2+y^2=356/2=178

x+y=16 x-y=10 So, x+y+x-y=16+10 So, 2x=26, x+13 So, y=16-13=3

Ans 13^2+3^2 =169+9 =178

The value of x is 13 and value of y is 3.Therefore,13^2+3^2=178.

add the 2 eqns to get x=13 put x=13 in any one eqn n get y=3 x x=169 y y=9 169+9=178