Algebra Know-It-ALL

330 Review Questions and Answers

To solve this, we add 15 to each side and then divide through by −5, as follows:

− 5 x− 15 = 0

− 5 x= 15

x=− 3

Question 16-9

How can we add multiples of the two original equations stated in Question 16-2 to solve the

linear system for y?

Answer 16-9

We can multiply the second equation through by −2 and then add it to the first equation,

getting the sum

2 x−y+ 8 = 0

− 2 x+ 6 y− 18 = 0

5 y− 10 = 0

To solve this, we add 10 to each side and then divide through by 5, like this:

5 y− 10 = 0

5 y= 10

y= 2

Question 16-10

How can we solve the two-by-two linear system stated in Question 16-2 by the rename-and-

replace (substitution) method? Here are the original equations again, for reference:

2 x−y+ 8 = 0

and

x− 3 y+ 9 = 0

Answer 16-10

We start by converting either of the two equations to SI form, so one of the variables appears

all alone on the left side of the equals sign. Let’s use the first equation and isolate y on the left

side. To manipulate the equation, we proceed just as we did in Answer 16-2:

2 x−y+ 8 = 0

−y+ 8 =− 2 x

−y=− 2 x− 8

y= 2 x+ 8