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/calculatorhttps://www.desmos.com/calculator

Hope this helped!