Introductory Algebra for College Students (7th Edition)

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

Chapter 4 - Review Exercises - Page 338: 30


The solution is $(0, 0)$.

Work Step by Step

To solve this system of equations, we want to rewrite the equations so that the coefficients for one of the variables differ only in sign. In this case, we will change the coefficients of the $x$ term so that these can be cancelled out. We want to find the least common multiple of the coefficients of the $x$ term, which is $14$, in this case. So we will multiply each equation by a number that will make the coefficient of the $x$ term $14$. We multiply the first equation by $7$ and the second equation by $2$ to get: $$-7(2x + 7y) = -7(0)$$ $$ 2(7x + 2y) = 2(0)$$ Use distributive property: $$-14x - 49y = 0$$ $$ 14x + 4y = 0$$ We can now cancel out the $x$ term by adding the two equations together: $$-45y = 0$$ We divide both sides by $-45$ to solve for $y$: $$y = 0$$ Now that we have a value for $y$, we can solve for $x$ by plugging the $y$ value into one of the equations: $$7x + 2(0) = 0$$ Multiply, according to order of operations: $$7x + 0 = 0$$ $$7x = 0$$ Divide both sides by $7$: $$x = 0$$ The solution is $(0, 0)$.
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