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Some Solutions to Homework #4
3.1 #60 Determine whether each of the ordered pairs a) (0,6), b) (−4,2), c) (−1,3), d) (7,6) is
a solution of the equation y 2 − 4x = 8:
Recall that the first number in an ordered pair represents the x-coordinate, and the second represents the y-coordinate. The ordered pair is a solution if when we evaluate both sides at the
prescribed values, we indeed have equality.
?
a) (6)2 − 4(0) = 8
×
36 − 0 = 8
This is not a solution.
?
b) (2)2 − 4(−4) = 8
×
4 + 16 = 8
This is not a solution.
?
c)(3)2 − 4(−1) = 8
×
9+4=8
This is not a solution.
?
d)(6)2 −√4(7) = 8
36 − 28 = 8
This is a solution.
3.2 #37: Find the x- and y-intercepts (if any) of the equation 4x − y + 3 = 0: To find where
the graph of an equation crosses the x-axis, use the fact that the x-axis is the line where y = 0.
Set y = 0 in the original equation and solve for x:
4x − (0) + 3 = 0
4x + 3 = 0
4x = −3
x = −3
4
So the x-intercept is at x = −3
, or the point ( −3
,0). To find the y-intercept, use the fact that the
4
4
y-axis is the line where x = 0. Set x = 0 in the original equation and solve for y:
4(0) − y + 3 = 0
−y + 3 = 0
−y = −3
y=3
So the y-intercept is at y = 3, or the point (0,3).
3.3 #14: Plot the points (7,1) and (4,−5) and find the slope of the line passing through them.
State whether this line is rising, falling, horizontal, or vertical:
1
(a good way the reI’ll skip the plotting part. To find the slope, use the formula m = xy22 −y
−x1
rise
member this formula is think of this formula as an algebraic incarnation of run : the difference of
y-coordinates is the “rise” and the difference in x-coordinates is the “run”). Here we have x1 = 7,
y1 = 1, x2 = 4, and y2 = −5. Then,
m = (−5)−(1)
(4)−(7)
m = −6
−3
We can simplify this fraction by cancelling a factor of 3 from the numerator and the denominator:
m = −2
=2
−1
So the slope is 2. Since the slope is positive, this line is rising (we can also see this from plotting
the two points).
3.3 #63: Write the equation of the line x − 4y + 2 = 0 in slope-intercept form, and use this
information to graph the line:
To put the equation of a line in slope-intercept form, solve for y:
x − 4y + 2 = 0
−4y + 2 = −x
−4y = −x − 2
(−x − 2)
y = −1
4
−1
y = 4 (−x) + −1
(−2)
4
y = 14 x + 12
With the equation in slope-intercept form, we see that the slope is 41 and the y-intercept is 12 . To
sketch a graph, plot the y-intercept point (0, 12 ). Then, since the slope is 14 , if we go over by 4 and
up by 1, we will have another point (4,1 12 ) on the line.
3.4 #56: Write the slope-intercept form of the equation that passes through the two points
(0,10) and (5,0):
Given two points, we can calculate the slope of the line between them as in 3.3 #14 above. We
will then have all the information required to use the point-slope formula.
m = 0−10
5−0
m = −10
5
m = −2
Then we can use the slope m = −2 and point (0,10) in the point-slope form y − y1 = m(x − x1 ):
y − 10 = −2(x − 0)
To rewrite this in the slope-intercept form, solve for y and simplify:
y − 10 = −2(x − 0)
y = −2(x − 0) + 10
y = −2x + 10
Although we knew two points, we had to choose just one to plug into the point-slope form. It
doesn’t matter which you choose. Let’s check that we would get the same slope-intercept form if
we had chosen the other point:
y − 0 = −2(x − 5)
y = −2(x − 5)
y = −2x − 2(−5)
y = −2x + 10
3.4#70 Write equations of the lines passing through (−5,−10) and are a) parallel b) perpendicular to the line 2x + 5y − 12 = 0.
The criteria for lines being parallel or perpendicular are in terms of the slopes of the lines. Let’s
start by finding the slope of the given line. We put the line in its slope-intercept form:
2x + 5y − 12 = 0
5y − 12 = −2x
5y = −2x + 12
+ 12
y = −2
5
5
The slope of the given line is −2
.
5
a) The slopes of parallel lines are equal. So, we want a line with slope −2
passing through the
5
point (−5,−10). The point-slope form quickly gives us the equation of such a line: y − (−10) =
−2
(x − (−5)).
5
−1
= 25
. So we want a line with slope ( −1
b) The slopes of perpendicular lines satisfy m2 = m
−2
)
1
5
passing through the point (−5,−10). Use the point-slope form again: y − (−10) = 52 (x − (−5)).
3.5 #29: Graph the solution set of the inequality x − 2y ≥ 6:
First, graph the boundary line x − 2y = 6 (see 3.3 #63 above). Graph this with a solid line since
we have an inclusive “or-equal-to” inequality.
x − 2y = 6
−2y = −x + 6
−x + 6)
y = −1
2
y = 12 x − 3
Then choose a test point not on the line to see which side to include. I’ll test (0,0):
?
0 − 2(0) ≥ 6
×
0≥6
The point (0,0) is not in the solution set, so the side we include is the side to the bottom right of
the boundary line: