An algebra problem by Yitong Yab

$\Rightarrow (x+y)^2 + (x-y)^2 = 18 + 12$

$\Rightarrow x^2+2xy+y^2 + x^2-2xy+y^2 = 30$

$\Rightarrow 2x^2+2y^2 = 30$

$\Rightarrow 2(x^2+y^2) = 30$

$\Rightarrow x^2+y^2 = \frac {30}{2}$

$\Rightarrow x^2+y^2 = 15$

Expanding (x+y)^2=18 and (x-y)^2=12, we get

x^2+Y^2+2xy=18

x^2+y^2-2xy=12

Add the above two equations

2(x^2+y^2)=30

So

x^2+y^2=15

>>answer is 15

$x ^ { 2 } + y ^ { 2 } = ( x \pm y ) ^ { 2 } \mp 2xy$ .

$x ^ { 2 } + 2xy + y ^ { 2 } = 18, x ^ { 2 } - 2xy + y ^ { 2 } = 12 \rightarrow 4xy = 6$

$\displaystyle \therefore xy = \frac { 3 } { 2 }$ .

$x ^ { 2 } + y ^ { 2 } = 18 - 3 = 15, 12 + 3 = 15$ .

${ ({ x }+y) }^{ 2 }=18...............(i)\\ { (x-y) }^{ 2 }=12...............(ii)\\ Expanding\quad (i)\quad \& \quad (ii),\quad we\quad have\\ { x }^{ 2 }+{ y }^{ 2 }+2xy=18..........(A)\quad and\\ { x }^{ 2 }+{ y }^{ 2 }-2xy=12..........(B)\\ Adding\quad (A)\quad \& \quad (B),\quad we\quad have\\ =>2({ x }^{ 2 }+{ y }^{ 2 })=30\\ =>{ x }^{ 2 }+{ y }^{ 2 }=30/2\\ =>\boxed { { x }^{ 2 }+{ y }^{ 2 }=15 }$