Exercise 20 Page 165

First, notice that the graph of the equation is a plane. Recall that two lines determine a plane. Therefore we need to find two distinct lines that lie on our plane in order to graph it. -3x+5y+10z=15 The lines contained in the plane can be found by substituting 0 for one of its variables. The resulting equation will represent a line. Let's first find the equation of the line that passes through x= 0.

Now that we know the equation of the line, we can use its intercepts to graph it. For the y-intercept, we will substitute z= 0. Similarly, for the z-intercept, we will substitute y= 0.

	y-intercept	z-intercept
Substitute	5y+10( 0)=15	5( 0)+10z=15
Calculate	y= 3	z= 1.5
Point	( 0, 3, 0)	( 0, 0, 1.5)

Next, we can plot the intercepts on the coordinate space and draw the line through them.

Now we will find the equation of another line on the plane. This time, let's look for the line that passes through y= 0.

-3x+5y+10z=15

Substitute

y= 0

-3x+5( 0)+10z=15

ZeroPropMult

Zero Property of Multiplication

- 3+0+10z=15

IdPropAdd

Identity Property of Addition

-3x+10z=15

To graph the line, we will find its intercepts by substituting z= 0 and x= 0.

	x-intercept	z-intercept
Substitute	-3x+10( 0)=15	-3( 0)+10z=15
Calculate	x= - 5	z= 1.5
Point	( - 5, 0, 0)	( 0, 0, 1.5)

Now that we know the intercepts, let's graph the second line!

Now we can graph both lines and the plane they determine.