College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 5 - Section 5.1 - Systems of Linear Equations - 5.1 Exercises - Page 487: 37

Answer

The system of equations is dependent so it has infinitely many solutions. The solution set is: $\left\{\left(\dfrac{6-2y}{7}, y\right)\right\}$.

Work Step by Step

We need to solve the given system of equations: $$7x+2y=6 ~~~(1) \\ 14x+4y=12 ~~~(2)$$ Multiply first equation by $-2$ and then first equation is equivalent to: $-14x-4y=-12 ~~~~(3)$ Add equations (2) and (3) to eliminate $x$ . $$14x+4y+(-14x-4y)=12+(-12) \\ 0=0 $$ Both variables were eliminated and resulted in a true equation/statement. This implies that the system of equations is dependent and has infinitely many solutions, and that the equatoins refer to the same line. Isolate $x$ from first equation to obtain: $$7x=6-2y\\ x=\dfrac{6-2y}{7}$$ Thus, the solution set is: $\left\{\left(\dfrac{6-2y}{7}, y\right)\right\}$.
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