## Elementary Algebra

Published by Cengage Learning

# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2: 18

#### Answer

A binder costs 2.29, a ream of paper costs 4.29, and a notebook costs .79 dollars.

#### Work Step by Step

We first set up three equations, using z as the cost of a ream of paper, y as the cost of a notebook, and x as the cost of a binder. $3x + z + 4y = 14.32 \\ x + 2z + 5y = 14.82 \\ 2x + 3y + 3z = 19.82$ In order to solve systems of three linear equations, we multiply the first and second equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by -2 and the second equation by 1 and add to obtain: $-5x -3y = -13.82$ We now multiply the first and third equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by -3 and the third equation by one and add to obtain: $-7x -9y = -23.14$ Plugging $x =-.6y +2.764$ into this equation, we obtain: $-4.8y -19.348 = -23.14 \\ y =.79$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = 2.29$ Finally, we plug the values of x and y into the first equation listed in the book to find: $z = 4.29$

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