   # For Problems 1-4, refer to the following systems of equations: I . 5 x - 2 y = 12 2 x + 5 y = 7 II . x - 4 y = 1 2 x - 8 y = 2 III . 4 x - 5 y = 6 4 x - 5 y = 1 IV . 2 x + 3 y = 9 7 x - 4 y = 9 For which of these systems are the equations said to be dependent? ### Intermediate Algebra

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

#### Solutions

Chapter
Section ### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
Publisher: Cengage Learning
ISBN: 9781285195728
Chapter 10.CT, Problem 1CT
Textbook Problem
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## For Problems 1-4, refer to the following systems of equations: I. 5 x - 2 y = 12 2 x + 5 y = 7 II. x - 4 y = 1 2 x - 8 y = 2 III. 4 x - 5 y = 6 4 x - 5 y = 1 IV. 2 x + 3 y = 9 7 x - 4 y = 9 For which of these systems are the equations said to be dependent?

To determine

To find:

The dependent system of equations.

### Explanation of Solution

Graphing a system of two linear equation in two variables produces one of the following results.

1. The graphs of two equations are two intersecting lines, which indicates that there are one unique solution of the system. Such a system is called consistent system.

2. The graphs of the two equations are parallel lines, which indicate that there is no solution for the system. It is called an inconsistent system.

3. The graph of two equations are the same line, which indicates infinitely many solutions for the system. The equations are called dependent equations.

Calculation:

(I)

5x-2y=12                 (1)2x+5y=7                   (2)

Let graph the lines by determining intercepts and a check point for each of the lines.

Solve the equation (1) for y we get,

5x-2y=12

Adding -5x on both sides we get,

5x-2y-5x=12-5x

-2y=12-5x

y=-12-5x2

 x y=-12-5x2 -2 -11 -1 -8.5 0 -6 2 -1

Solve the equation (2) for y we get,

2x+5y=7

Adding -2x on both sides we get,

2x+5y-2x=7-2x

5y=7-2x

y=7-2x5

 x y=7-2x5 -2 2.2 -1 1.8 0 1.4 1 1

The graph of the given system is,

(II)

x-4y=1               (1)2x-8y=2                  (2)

Let graph the lines by determining intercepts and a check point for each of the lines.

Solve the equation (1) for y we get,

x-4y=1

Adding x on both sides we get,

x-4y-x=1-x

-4y=1-x

y=x-14

 x y=-(1-x)4 -1 -0.5 0 -0.25 1 0 2 0.25

Solve the equation (2) for y we get,

2x-8y=2

Adding 2x on both sides we get,

2x-8y-2x=2-2x

-8y=2-2x

y=2x-28

y=2x-18

y=2x-18

 x y=-(1-x)4 -1 -0.5 0 -0.25 1 0 2 0

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