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 - Section 4.3 - Solving Systems of Linear Equations by the Addition Method - Exercise Set - Page 310: 14


The solution is $(2, 1)$.

Work Step by Step

We first need to modify the two equations so that the coefficients of one of the variables differs only in sign. With this modification, we will be able to cancel out one of the variables and solve for the other one: Let us modify the equations first. We see that if we multiply the second equation by $5$ and leave the first equation alone, the coefficients of the $y$ term will be $-5$ and $5$. The coefficients will now only differ in sign, so we can cancel them out when we add the two equations. Let us do the multiplication: $2x - 5y = -1$ $5(3x + y) = 5(7)$ Use distributive property: $2x - 5y = -1$ $5(3x) + 5(y) = 5(7)$ Let's multiply out the terms: $2x - 5y = -1$ $15x + 5y = 35$ We can now cancel out the $y$ terms to get: $2x = -1$ $15x = 35$ Now we add both sides of the two equations to get: $17x = 34$ Divide both sides of the equation by $17$ to solve for $x$: $x = 2$ Now that we have the value for $x$, we can plug this value into one of the equations to solve for $y$. Let's use the second equation: $3(2) + y = 7$ Multiply first: $6 + y = 7$ Subtract $6$ from both sides of the equation to solve for $y$: $y = 1$ The solution is $(2, 1)$.
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