Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Cumulative Review Exercises - Page 685: 5

Answer

The solution is $(3, 4)$.

Work Step by Step

We can use the substitution method for this problem because in the second equation, $y$ is already isolated. We can use the expression for $y$ to substitute into the first equation: $$3x - 2(10 - 2x) = 1$$ Use the distributive property to simplify: $$3x - (2)(10) - (2)(-2x) = 1$$ Multiply out the terms: $$3x - 20 + 4x = 1$$ Group like terms: $$(3x + 4x) - 20 = 1$$ Combine like terms: $$7x - 20 = 1$$ Add $20$ to each side to isolate the variable to one side of the equation and the constant to the other: $$7x = 21$$ Divide each side of the equation by $7$ to solve for $x$: $$x = 3$$ Now that we have the value for $x$, we can substitute this value into the second equation to find $y$: $$y = 10 - 2(3)$$ Divide first, according to order of operations: $$y = 10 - 6$$ Subtract to solve for $y$: $$y = 4$$ The solution is $(3, 4)$.
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