Exercise 65 Page 650

We can solve the given system of linear equations using the Elimination Method. To do it, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This means that either the x-terms or the y-terms must cancel each other out. 5 x^2+ y^2=30 & (I) 9 x^2- y^2=-16 & (II) We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).

Now we can solve for y by substituting each value of x into Equation (I) and simplifying. Let's start with x=1.

This gave us two solutions, or points of intersection, of the system of equations, (1, 5) and (1,- 5). Let's now substitute - 1 for x in Equation (I).

We have now found another two solutions of the system, (- 1, 5) and (- 1,- 5). Thus, the given system has four solutions, (1, 5), (1,- 5), (- 1, 5), and (- 1,- 5). We can also state them as (1, ±5) and (-1, ±5).