Download MAT096 Lab Sheet 4 - LaGuardia Community College

LAGUARDIA COMMUNITY COLLEGE
Department of Mathematics, Engineering, and Computer Science
MAT96 ELEMENTARY ALGEBRA
ALEKS PILOT
LAB # 4
Name: ____________________________________
Date: ____________________
Instructor: _________________________________
Section: __________________
You need to show all work. Indicate the right answer in the answer sheet. Even if you mark the
right answer, but do not show work on this sheet, you will not be given credit for that question:
Tutor problems are the examples for the unsolved problem(s) that follow them.
1. (Tutor) Check which two of the following are solutions to the equation 2x – 5y = 7.
a) (0, –1)
b) (1, 1)
c) (1, – 1)
d) (3,
)
e) (6, −1)
Solution: We substitute each ordered pair in the equation and check if the right side equals the
left side.
Check (a): We substitute (0, -1) for (x, y) in the left side of the equation and check if we get 7.
We get 2(0) – 5(-1) = 0 + 5 = 5 ≠ 7. So, (0, -1) is not a solution to the equation 2x – 5y = 7.
Check (b): We substitute (1, 1) for (x, y) in the left side of the equation and check if we get 7.
We get 2(1) – 5(1) = 2 – 5= -3 ≠ 7. So, (1, 1) is not a solution to the equation 2x – 5y = 7.
Check (c): We substitute (1, -1) for (x, y) in the left side of the equation and check if we get 7.
We get 2(1) – 5(-1) = 2 + 5 = 7. So, (1, -1) is a solution to the equation 2x – 5y = 7.
Check (d): We substitute (3,
We get
= 7.
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) for (x, y) in the left side of the equation and check if we get 7.
. So, (3,
) is a solution to the equation 2x – 5y
Check (e): We substitute (6, -1) for (x, y) in the left side of the equation and check if we get 7.
We get 2(6) – 5(-1) = 12 + 5= 17 ≠ 7. So, (6, -1) is not a solution to the equation 2x – 5y = 7.
2. Check which two of the following are solutions to the equation x + 2y = 2.
a) (6, 2)
b) (6, -2)
c) (1, -2)
d) (1, 1/2)
e) (1, −1/2)
3.(Tutor) Draw the graph of the equation
plotting three points
x
?
3
-5
by completing the table below and
y
-1
?
?
Solution: For the first point, it is given that y= -1 , so substituting y= -1 we get
; So the first point on the line is (-1, -1)
For the second point, we substitute x = 3 in the give equation
; Therefore, the second point is (3, 2)
Similarly, we can easily find the third point as (-5, -4)
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Now plotting these three points (-1, -1), ( 3, 2), and ( -5, -4) we can graph a line as shown above
4.Draw the graph of the equation
three points
x
?
4
0
by completing the table below and plotting
y
2
?
?
5.(Tutor) The slope of a straight line through the points (3, 8) and (4, 5) is:
a) 1/3
b) 3
c) – 1/3
d) – 3
e) 0
Solution: Slope (m) of a line passing through two points (
the formula:
.
Slope of a line passing through (3, 8) and (4, 5) is:
=
and (
= -3. So, the answer is d).
6. The slope of a straight line through the points (– 1, –3) and (2, 5) is:
a) 8
1
b)
2
3
c)
5
d) 3/8
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can be found using
e)
8
3
7. (Tutor) The slope of the line 4x – 2y = 3 is:
a) – 4
b)
4
c)
3
2
d)
1
2
e) 2
Solution: When the equation of a line is given, slope of the line can be found by rewriting the
equation in the y = mx + b form where ‘m’ the co-efficient of x, is the slope of the line. So, we
first rewrite the given equation 4x – 2y = 3 in the y = mx + b form (i.e. keep ‘y’ by itself on the
left hand side).
So, we solve the equation for ‘y’ as: - 2y = - 4x + 3
Divide both sides by -2, we get
After simplifying, we get
Slope is 2. So, the answer is e).
8. The slope of the line 2x – 5y = 3 is:
a) 2
b) – 5
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.
c) 5/2
d) 2/5
e) –3
9. (Tutor) Draw graph for the equation y + 3x = 5 using slope and intercept values.
Solution: Let’s find slope and y-int of the equation y + 3x = 5:
Rewrite the equation as: y = -3x +5
Slope: m=-3=-3/1 , b = 5, yint = (0,5)
We plot the point (0,5) and to find second point we use the slope -3/1. We move 3 units down
and 1 to the right from (0,5)
y
x
10..Draw graph for the equation 2x + 3y = 12 using slope and intercept values.
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y
x
11. (Tutor) Find the equation of the line with slope -3 and passing through a point (-2, 5)
Solution: The equation of the line in slope intercept form is
Since the slope of a line is -3, the equation must be
To determine value of b we now can use the fact that the line passes through (-2, 5).
Substituting
we have,
Therefore
12. Find the equation of the line with slope -5 and passing through a point (1, -3).
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13. (Tutor) Determine the equation of the line passing through the points (−3, 4) and (2, 1)
a.
b.
c.
d.
e.
Solution: The line passes through (-3, 4) and (2, 1)
Slope (m) =
We will use the point (-3, 4) and (2,1) and the slope
to find the equation
: Add 4 on both sides
+4
+4
We now write this equation in standard form.
5
: Multiply both sides by 5
: Add
on both sides
14. Determine the equation of the line passing through the points (3, 8) and (− 6, 2).
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Extra Practice Problems (Optional)
1. The slope of a straight line through the points (– 4,1) and (–2,3) is:
a) (-1)/3
b) 5
c) -1
d) 1
e) 1/3
2. The slope of the line is:  4 x  3 y  12
a) 4
b) -4
c) (-4)/3
d) 4/3
e) 3
3. Find the equation of the straight line with slope 3/2 passing through a point (-2,0)
4. Determine the equation of the line passing through the points (0, 3) and (-1, 4).
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