Exercise 9 Page 269

In order to find which systems are equivalent, let's reduce the systems as much as possible by removing common factors from the equations. Then we can compare them more easily. Let's take a look into the first and third systems. 4x-5y=3 2x+15y=-1 and 4x-5y=3 4x+30y=-1 Notice that these sytems are already as simplified as possible because the equations in them do not have any common multiples. Now, let's consider the second system. Equation (II) in the second system can be reduced. It has a common multiple of 2, which means we can divide the entire equation by 2. 4x-5y=3 & (I) -4x -30y= 2 & (II) ⇒ 4x-5y=3 -2x -15y= 1 Last, the fourth system can also be reduced. Since it has a common multiple of 3 in the first equation, we can divide the entire equation by 3. 12x -15y= 9 & (I) 2x+15y=-1 & (II) ⇒ 4x -5y= 3 2x+15y=-1 Now we can compare the reduced systems side by side.

When two systems of equations are multiples of one another, they are equivalent and will have the same solution. Therefore, as we can see from the table, the first and fourth systems are equivalent.