Algebra Know-It-ALL

Part Two 331

Next, we substitute the quantity (2x+ 8) for y in the second equation and solve the result for
x, as follows:

x− 3 y+ 9 = 0

x− 3(2x+ 8) + 9 = 0

x− 6 x− 24 + 9 = 0

− 5 x− 15 = 0

− 5 x= 15

x=− 3

Now that we know the value of x, we can plug it into either equation and solve for y. Let’s use
the first equation. Then we proceed as follows:

2 x−y+ 8 = 0

2 × (−3)−y+ 8 = 0

− 6 −y+ 8 = 0

− 6 + 8 =y

y= 2

Chapter 17

Question 17-1

Let’s consider again the two-by-two system we saw in Question 16-2. How can we graph this
system in Cartesian coordinates, with x as the independent variable and y as the dependent
variable? Here are the original equations:

2 x−y+ 8 = 0

and

x− 3 y+ 9 = 0

Answer 17-1

We can use the SI forms of the equations to find their y-intercepts, and the solution of the sys-
tem to find a third point that lies on both lines. We’re lucky here, because the intersection point
is fairly far away from the y axis. The SI forms of the equations were derived in Answer 16-2.
Respectively, they are:

y= 2 x+ 8