1 y 1 2 x 1 y 2 6 x 3 2 y 1 3 x 6 y 2 3 x 2 3 y 1 x 3 y 2 2 x 12 may have one

# 1 y 1 2 x 1 y 2 6 x 3 2 y 1 3 x 6 y 2 3 x 2 3 y 1 x 3

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1. y 1 = 2 x + 1 y 2 = 6 x + 3 2. y 1 = 3 x + 6 y 2 = 3 x + 2 3. y 1 = x 3 y 2 = 2 x + 12 may have one solution, no solution, or an infinite number of solutions. Page 11 of 74 Print | Beginning & Intermediate Algebra 1/31/2014 ...
PRINTED BY: [email protected] Printing is for personal, private use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Case I: One solution . The two graphs intersect at one point, which is the solution. We say that the equations are independent. It is a consistent system of equations. There is a point (an ordered pair) consistent with both equations. Case II: No solution . The two graphs are parallel and so do not intersect. We say that the system of equations is inconsistent because there is no point consistent with both equations. Page 12 of 74 Print | Beginning & Intermediate Algebra 1/31/2014 ...
PRINTED BY: [email protected] Printing is for personal, private use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Case III: An infinite number of solutions . The graphs of each equation yield the same line. Every ordered pair on this line is a solution to both of the equations. We say that the equations are dependent. EXAMPLE 7 If possible, solve the system. 2 x + 8 y = 16 (1) 4 x + 16 y = 8 (2) Solution To eliminate the variable y, we’ll multiply equation (1) by 2. 2(2 x ) + ( 2)(8 y ) = ( 2)(16) 4 x 16 y = 32 (3) We now have the following equivalent system. 4 x 16 y = 32 (3) 4 x + 16 y = 8 (2) When we add equations (3) and (2) , we get 0 = 40, which, of course, is false. Thus, we conclude that this system of equations is inconsistent, and there is no solution. Therefore, equations (1) and (2) do not intersect, as we can see on the graph to the right. If we had used the substitution method to solve this system, we still would have obtained a false statement. When you try to solve an inconsistent system of linear equations by any method, you will always obtain a mathematical equation that is not true. Student Practice 7 If possible, solve the system. 4 x 2 y = 6 6 x + 3 y = 9 Page 13 of 74 Print | Beginning & Intermediate Algebra 1/31/2014 ...
PRINTED BY: [email protected] Printing is for personal, private use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. 281282 EXAMPLE 8 If possible, solve the system. 0.5 x 0.2 y = 1.3 (1) 1.0 x + 0.4 y = 2.6 (2) Solution Although we could work directly with the decimals, it is easier to multiply each equation by the appropriate power of 10 (10, 100, and so on) so that the coefficients of the new system are integers. Therefore, we will multiply equations (1) and (2) by 10 to obtain the following equivalent system.

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