Substitute Them Wisely

Algebra Level 3

{ y + 6 = ( x − 3 ) 2 x + 6 = ( y − 3 ) 2 \large{ \begin{cases} y+6 = (x-3)^2 \\ x+6 = (y-3)^2 \end{cases}}

If x x and y y are distinct numbers that satisfy the system of equations above, find x 2 + y 2 x^2+y^2 .


The answer is 29.

View wiki
Choi Chakfung by choi chakfung , 28, Hong Kong

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ikkyu San
Mar 20, 2016

Let

y + 6 = ( x − 3 ) 2 ⇒ ( 1 ) x + 6 = ( y − 3 ) 2 ⇒ ( 2 ) \begin{aligned}\color{#D61F06}{y+6=}&\ \color{#D61F06}{(x-3)^2\Rightarrow(1)}\\\color{#3D99F6}{x+6=}&\ \color{#3D99F6}{(y-3)^2\Rightarrow(2)}\end{aligned}

Equation ( 1 ) − \color{#D61F06}{(1)}- Equation ( 2 ) \color{#3D99F6}{(2)} ;

( y + 6 ) − ( x + 6 ) = ( x − 3 ) 2 − ( y − 3 ) 2 y + 6 − x − 6 = [ ( x − 3 ) − ( y − 3 ) ] [ ( x − 3 ) + ( y − 3 ) ] y − x = ( x − 3 − y + 3 ) ( x − 3 + y − 3 ) − ( x − y ) = ( x − y ) ( x + y − 6 ) − 1 = x + y − 6 x + y = 5 x 2 + 2 x y + y 2 = 25 x 2 + y 2 = 25 − 2 x y ⇒ ( 3 ) \begin{aligned}\color{#D61F06}{(y+6)}-\color{#3D99F6}{(x+6)}=&\ \color{#D61F06}{(x-3)^2}-\color{#3D99F6}{(y-3)^2}\\y+6-x-6=&\ [(x-3)-(y-3)][(x-3)+(y-3)]\\y-x=&\ (x-3-y+3)(x-3+y-3)\\-(x-y)=&\ (x-y)(x+y-6)\\\color{#302B94}{-1=}&\ \color{#302B94}{x+y-6}\\\color{#20A900}{x+y=}&\ \color{#20A900}5\\x^2+2xy+y^2=&\ 25\\x^2+y^2=&\ 25-2\color{#624F41}{xy}\Rightarrow(3)\end{aligned}

Equation ( 1 ) + \color{#D61F06}{(1)}+ Equation ( 2 ) \color{#3D99F6}{(2)} ;

( y + 6 ) + ( x + 6 ) = ( x − 3 ) 2 + ( y − 3 ) 2 y + 6 + x + 6 = [ ( x − 3 ) + ( y − 3 ) ] 2 − 2 ( x − 3 ) ( y − 3 ) x + y + 12 = ( x + y − 6 ) 2 − 2 ( x y − 3 x − 3 y + 9 ) 5 + 12 = ( − 1 ) 2 − 2 [ x y − 3 ( x + y ) + 9 ] 17 = 1 − 2 [ x y − 3 ( 5 ) + 9 ] 16 = − 2 ( x y − 15 + 9 ) − 8 = x y − 6 x y = − 2 \begin{aligned}\color{#D61F06}{(y+6)}+\color{#3D99F6}{(x+6)}=&\ \color{#D61F06}{(x-3)^2}+\color{#3D99F6}{(y-3)^2}\\y+6+x+6=&\ [(x-3)+(y-3)]^2-2(x-3)(y-3)\\\color{#20A900}{x+y}+12=&\ (\color{#302B94}{x+y-6})^2-2(\color{#624F41}{xy}-3x-3y+9)\\\color{#20A900}5+12=&\ (\color{#302B94}{-1})^2-2[\color{#624F41}{xy}-3(\color{#20A900}{x+y})+9]\\17=&\ 1-2[\color{#624F41}{xy}-3(\color{#20A900}5)+9]\\16=&\ -2(\color{#624F41}{xy}-15+9)\\-8=&\ \color{#624F41}{xy}-6\\\color{#624F41}{xy=}&\ \color{#624F41}{-2}\end{aligned}

Instead x y \color{#624F41}{xy} with − 2 \color{#624F41}{-2} in Equation ( 3 ) (3) ;

x 2 + y 2 = 25 − 2 ( − 2 ) = 25 + 4 = 29 x^2+y^2=25-2(\color{#624F41}{-2})=25+4=\boxed{29}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Very tedious, just find out x+y like you did then add both equations you will find that only x^2+y^2=29

Kushagra Sahni - 5 years, 2 months ago

y+6=(x-3) 2 ^{2} (eqn1)
x+6=(y-3) 2 ^{2} (eqn2)
eqn1+eqn2=x+y+12=x 2 ^{2} -6x+9+y 2 ^{2} -6y+9
x 2 ^{2} +y 2 ^{2} =7x+7y-6
x 2 ^{2} +y 2 ^{2} =7(x+y)-6
eqn1-eqn2=y-x=x 2 ^{2} -y 2 ^{2} -6x+6y
5x-5y=x 2 ^{2} -y 2 ^{2}
5(x-y)=(x+y)(x-y)
5=x+y
x 2 ^{2} +y 2 ^{2} =7(x+y)-6
x 2 ^{2} +y 2 ^{2} =7(5)-6
x 2 ^{2} +y 2 ^{2} =29



0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...