# Use a graphing calculator to check your answers for Problem 65. 65. A system such as ( 3 x + 2 y = 2 2 x − 3 y = 1 4 ) is not a system of linear equations nut can be transformed into a linear system by changing variables. For example, when we substitute u for 1 x and v for 1 y , the system cited becomes ( 3 u + 2 v = 2 2 u − 3 v = 1 4 ) We can solve this “new” system either by elimination by addition or by substitution (we will leave the details for you) to produce u = 1 2 and v = 1 4 . Therefore, because u = 1 x and v = 1 y , we have 1 x = 1 2 and 1 y = 1 4 Solving these equations yields, x = 2 and y = 4 The solution set of the original system is { ( 2 , 4 ) } . Solve each of the following system. (a) ( 1 x + 2 y = 7 12 3 x − 2 y = 5 12 ) (b) ( 2 x + 3 y = 19 15 − 2 x + 1 y = − 7 15 ) (c) ( 3 x − 2 y = 13 6 2 x + 3 y = 0 ) (d) ( 4 x + 1 y = 11 3 x − 5 y = − 9 ) (e) ( 5 x − 2 y = 23 4 x + 3 y = 23 2 ) (f) ( 2 x − 7 y = 9 10 5 x + 4 y = − 41 20 ) BuyFind

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
Publisher: Cengage Learning
ISBN: 9781285195728 BuyFind

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
Publisher: Cengage Learning
ISBN: 9781285195728

#### Solutions

Chapter 10.3, Problem 67PS
Textbook Problem

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