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!