How do you solve for #x# and #y# in the simultaneous equations #x + 2y = 3# and #x^2 - y^2 = 24#?

1 Answer
Apr 18, 2018

The two solutions are #(-7,5)# and #(5,-1)#.

Explanation:

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):

https://www.desmos.com/calculator

Hope this helped!