Solutions of Equations

Algebra Level 2

x + y 2 = 2 x x + y = 2 y 2 \large \begin{aligned} x+y^2 & =2x \\ x+y & =2y^2 \end{aligned}

Solve the system of equations above.

( 5 , 2 ) (5,-2) and ( 2 , 3 ) (-2,3) ( 1 , 2 ) (1,2) and ( 0 , 1 ) (0,-1) ( 2 , 3 ) (2,3) and ( 5 , 5 ) (5,-5) ( 1 , 1 ) (1,1) and ( 0 , 0 ) (0,0)

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

Ryan Shi
Sep 13, 2016

Correct me if I'm wrong, but I just did:

x + y 2 = 2 x x + y^2 = 2x

y 2 = x y^2 = x ...... Equation 1

x + y = 2 y 2 x + y = 2y^2

x + y = 2 x x + y = 2x ...... From Equation 1

H e n c e Hence y = x y = x

Out of the four answers only ( 1 , 1 ) (1,1) and ( 0 , 0 ) (0,0) has property y = x y = x

Viki Zeta
Sep 12, 2016

x + y 2 = 2 x 2 x x = y 2 x = y 2 x + y = 2 y 2 y 2 + y = 2 y 2 y 2 = y y 2 y = 0 y ( y 1 ) = 0 y = 0, y = 1 y = 1 x = y 2 x = 0 y = 0 x = 1 2 = 1 x + y^2 = 2x \\ 2x - x = y^2 \\ x = y^2 \\ x + y = 2y^2 \\ y^2 + y = 2y^2 \\ y^2 = y \\ y^2 - y = 0 \\ y(y-1) = 0 \\ \fbox{ y = 0, y = 1 } \\ \fbox{ y = 1 } \\ \implies x = y^2 \implies x = 0 \\ \fbox{ y = 0 } \\ \implies x = 1^2 = 1

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...