How do you solve the system #x^2+y^2=25, x+2y=10#?
1 Answer
Mar 28, 2018
Explanation:
Given:
#{ (x^2+y^2=25), (x+2y=10) :}#
From the second equation, we have
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 = y^2-8y+20#
Subtract
#0 = y^2-8y+15 = (y-3)(y-5)#
So
If
#x = 10-2y = 10-6 = 4#
If
#x = 10-2y = 10-10 = 0#
So:
#(x, y) = (0, 5)" "# or#" "(x, y) = (4, 3)#