How do you solve the system #y = 2x -3#, #x^2 + y^2 = 2#?

Question

3488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-15T05:11:22

#(7/5, -1/5), (-1. -5)#

Replace #y# with its equivalent based from the linear equation and solve for #x#

#y = 2x - 3#

#x^2 + y^2 = 2#

#=> x^2 + (2x - 3)^2 = 2#

#=> x^2 + 4x^2 - 12x + 9 = 2#

#=> 5x^2 - 12x + 9 = 2#

#=> 5x^2 - 12x + 7 = 0#

#=> (5x - 7)(x - 1) = 0#

#=> x = 7/5, x = -1#

Now let's find the corresponding value of #y#

#x = 7/5#

#=> y = 2(7/5) - 3#
#=> y = 14/5 -3#

#=> y = (14 - 15)/5#

#=> y = -1/5#

#x = -1#

#=> y = 2(-1) - 3#

#=> y = -2 - 3#

#=> y = -5#