How do you solve the system #x^2+y^2=25, x+2y=10#?

Question

14512 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T21:48:47

#(x, y) = (0, 5)" "# or #" "(x, y) = (4, 3)#

Given:

#{ (x^2+y^2=25), (x+2y=10) :}#

From the second equation, we have #x = 10-2y#.

Substituting this in the first equation, we find:

#25 = x^2+y^2#

#color(white)(25) = (10-2y)^2+y^2#

#color(white)(25) = 5y^2-40y+100#

Divide both ends by #5# to get:

#5 = y^2-8y+20#

Subtract #5# from both sides to get:

#0 = y^2-8y+15 = (y-3)(y-5)#

So #y = 3# or #y = 5#

If #y=3# then:

#x = 10-2y = 10-6 = 4#

If #y = 5# then:

#x = 10-2y = 10-10 = 0#

So:

#(x, y) = (0, 5)" "# or #" "(x, y) = (4, 3)#