To solve the system of equations:
#{(x+2y=3,qquad(1)),(x^2-y^2=24,qquad(2)):}#
We must realize that there may be two solutions, since equation #(2)# is a hyperbola, and hyperbola may cross a linear equation at two points.
First, solve for #x# in equation #(1)#:
#x+2y=3#
#x=3-2y#
Plug this into equation #(2)#:
#x^2-y^2=24#
#(3-2y)^2-y^2=24#
#9-12y+4y^2-y^2=24#
#9-12y+3y^2=24#
#3-4y+y^2=8#
#y^2-4y-5=0#
#(y-5)(y+1)=0#
#y=-1,5#
Plug #y=-1# back into equation #(1)# and solve for its corresponding #x# value:
#x+2(-1)=3#
#x-2=3#
#x=5#
This means that the first solution is #x=5# and #y=-1#, or the point #(5,-1)#. Now plug #y=5# into equation #(1)#:
#x+2(5)=3#
#x+10=3#
#x=-7#
This means that the other solution is #x=-7# and #y=5#, or the point #(-7,5)#. Here is a graph of both of the functions (equation #(1)# is in red and equation #(2)# is in blue):
Hope this helped!
