## Elementary Algebra

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 1 and the second equation by 1 and add to obtain: $3x + 2y = 4$ 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 2 and the third equation by one and add to obtain: $5x + 7y =14$ Plugging $x = -2/3y + 4/3$ into this equation, we obtain: $2.67y + 6.67 = 14 \\ y =2$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = 0$ Finally, we plug the values of x and y into the first equation listed in the book to find: $z = 5$