How do you solve the simultaneous equations #x – 2y = 1# and #x^2 + y^2 = 29#?

1 Answer
Jul 28, 2015

Answer:

I found:
#x=5#; #y=2#
#x=-23/5#;#y=-15/5#

Explanation:

We can try isolate #x# from the first equation:
#x=1+2y#
substitute into the second (for #x#):
#(color(red)(1+2y))^2+y^2=29#
#1+4y+4y^2+y^2=29#
#5y^2+4y-28=0#
solve for #y# (using the Quadratic Formula) gives you:
#y_(1,2)=(-4+-sqrt(16+560))/10=(-4+-24)/10#
you get two solutions:
#y_1=-28/10=-14/5#
#y_2=20/10=2#
Substituted back to find #x# you get:
if #y=2# then #x=1+4=5#
if #y=-15/5# then #x=1-2*14/5=-23/5#