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