Download Math 99 – Homework 1

Math 99 – Test 1 Review Spring 2013
Math 99 – Test 1 Practice
1.
For the function f(x) = x2 – 2 x + 1 Find the following
(a) f(2)
(b) f(– 3)
(c) f( ½ )
2.
Solve the equation 3x + 7 =
28
3.
Solve the equation 5x + 7 =
– 2x + 35
4.
Solve the equation
3(2x – 2)
5.
Solve the equation
5
=
24
 10
x
6.
Solve the equation
x
4
2
=
7.
Solve the following Inequalities.
(a) 5x + 5 > 3x – 9
show all your working.
show all your working.
=
show all your working.
4x – 4
show all your working.
show all your working.
1
x2
3
show all your working.
(b) 2(x – 7) + 9 < 7x + 15
(c)
8.
Change the formula
y = mx + b to be in terms of m
9.
Peter’s height is 5 inches more than three times Mary’s height.
If Mary height is m inches write down Peters height in terms of m.
10. A bookcase is to have five shelves as shown in the diagram below.
The height of the bookcase is to be two feet more than its width.
Find the width and height of the bookcase if 50 feet of lumber
was used in the construction of the bookcase.
11. Express each of the system of equations below in the slope intercept form ( y = ….. form).
And without drawing graphs identify each system of linear equations below as consistent, inconsistent or
dependant. Also indicate the number of solutions that would occur.
(a)
y
2y
=
=
2x + 4
4 – 2x
(b) x – 2y
3x
=
=
10
6y + 15
(c)
x–4
6y + 8
=
=
3y
2x
Page 1
Math 99 – Test 1 Review Spring 2013
12.
Determine which of the following points, A(2,1) , B(1, – 3) and C(– 1 , 3) if any, satisfy both pairs of
equations.
y
3x – 2y
=
=
4x – 7
4
13. Graph the line 3x + 2y = 6 by plotting at least 3 points.
14. What is the slope and y-intercept of the line with equation 4x + 2y = 20
15.(a)
Graph the following lines.
2x + 3y
4x – 2y
=
=
15.(b)
From the graph what is the solution to this system of equations.
16.(a)
Solve the system of equations
y
y
16.(b)
Solve the system of equations
16.(c)
12
8
3x – 1
5x – 5
by using the substitution method.
2x + y
3x + 4y
= 1
= –1
by using the substitution method.
Solve the system of equations
5x – 2y
5
=
=
–7
y – 3x
by using the substitution method.
16.(d)
Solve the system of equations
4x – 2y
y
=
=
–7
2x – 1
by using the substitution method.
17.(a)
Solve the system of equations
5x – 2y
2x + 4y
=
=
8
8
by using the addition method.
17.(b)
Solve the system of equations
2x – 2y
4x + 3y
=
=
10
–1
by using the addition method.
17.(c)
Solve the system of equations
=
=
–1
y–2
by using the addition method.
4x
18.
=
=
A truck rental agency charges a daily fee plus a mileage fee. Julie was charged $85 for two days and 100 miles
and Christina was charged $165 for 3 days and 400 miles. What is the agency’s daily fee and what is the
mileage fee?
19. The plumber charges a fixed rate for turning up at your house plus a charge per hour for the work done. From
previous jobs that the plumber has done it was found that a 4 hour job will cost a total of $283 while a 6 hour
job will cost a total of $387.What is the fixed rate and the charge per hour?
20. By weight one alloy of brass is 70% Copper and 30% Zinc. Another Alloy is 40% Copper and 60% Zinc. How
many grams of each alloy would need to be melted and combined to obtain 600 grams of a brass alloy that is
60% Copper and 40% Zinc?
Page 2
Math 99 – Test 1 Review Spring 2013
Math 99 – Test 1 Practice Solutions
1.
For the function f(x) = x2 – 2 x + 1 Find the following
(a) f(2) = (2)2 – 2 (2) + 1 = 4 – 4 + 1 = 1
(b) f(– 3) = (– 3)2 – 2 (– 3) + 1 = 9 + 6 + 1 = 16
(c) f( ½ ) = (½)2 – 2 (½) + 1 = ¼ – 1 + 1 = ¼
2.
3.
Solve the equation 3x + 7 =
28
3x + 7
3x
x
subtract 21 from both sides
divide both sides by 3
=
=
=
=
– 2x + 35
+ 35
28
4
Solve the equation
3(2x – 2)
6x – 6
2x – 6
2x
x
5.
28
21
7
Solve the equation 5x + 7 =
5x + 7
7x + 7
7x
x
4.
=
=
=
=
=
=
=
=
 10
x
5x =
x =
– 240
– 48
– 2x + 35
show all your working.
add 2x to both sides
subtract 7 from both sides
divide both sides by 7
3(2x – 2)
=
4x – 4
4x – 4
–4
2
1
Solve the equation
5
=
24
show all your working.
5
=
24
4x – 4
show all your working.
multiply out 3(2x – 2)
subtract 4x from both sides
add 6 to both sides
divide both sides by 2
 10
x
show all your working.
cross multiply
divide both sides by 5
Page 3
Math 99 – Test 1 Review Spring 2013
6.
Solve the equation
x
4 =
2
1
x4 =
2
1
1
x x4
2
3
1
x4
6
1
x
6
x
7.
x
4 =
2
1
x2
3
1
x2
3
=
–2
=
–2
=
–6
=
– 36
1
x2
3
show all your working.
1
x
as being the same as x
2
2
1
subtract x from both sides
3
1
1
1
simplifying
x x = x
2
6
3
think of
subtract 4 from both sides
multiply both sides by 6 or divide both sides by
1
6
Solve the following Inequalities.
(a) 5x + 5
5x
2x
x
>
>
>
>
3x – 9
3x – 14
– 14
–7
(b) 2(x – 7) + 9
2x – 14 + 9
2x – 5
2x
– 5x
x
<
<
<
<
<
>
7x + 15
7x + 15
7x + 15
7x + 20
20
–4
(c)
( multiply both sides by 12 )
(divide both sides by – 3 and switch the inequality symbol )
8.
Change the formula
y
y–b
yb
x
9.
y = mx + b to be in terms of m
=
=
mx + b
mx
subtract b from both sides
=
m
divide both sides by x
Peter’s height is 5 inches more than three times Mary’s height.
If Mary height is m inches write down Peters height in terms of m.
Peter’s height
=
3 times Mary’s height + 5
=
3m + 5
Page 4
Math 99 – Test 1 Review Spring 2013
10. A bookcase is to have five shelves as shown in the diagram below.
The height of the bookcase is to be two feet more than its width.
Find the width and height of the bookcase if 50 feet of lumber was used in the construction
ww
of the bookcase.
Total amount of wood
w + 2 + w +w + w + w +w + w + 2
7w + 4
7w
50 feet
w+2
50
50
46
46
w
=
or 6.57 feet
7
width of bookshelf = w = 6.75 feet and height of bookshelf = 6.75 + 2
w
w
w
=
=
=
=
w+2
w
w
w
w
=
8.75 feet
11. Express each of the system of equations below in the slope intercept form ( y = ….. form).
And without drawing graphs identify each system of linear equations below as consistent, inconsistent or
dependant. Also indicate the number of solutions that would occur.
(b)
y
2y
= 2x + 4
=
4 – 2x
2y
y
=
=
4 – 2x
2–x
y
=
–x+2
The two equations are
Now
(b) x – 2y
3x
(c)
x–4
6y – 8
=
=
10
6y + 15
x – 2y
– 2y
=
=
10
– x + 10
x–4
= 3y
3y = x – 4
y
=
½x–5
y
3x
6y + 15
=
=
6y + 15
3x
6y + 8
6y
6y
=
3x – 15
y
y
=
½x – 2½
=
= 3y
= 2x
1
4
x
3
3
=
=
2x
2x
–8
y
y
=
=
2x + 4
–x+2
Consistent
# of solutions = 1
Inconsistent
# of solutions = 0
=
1
4
x
3
3
Dependant
# of solutions =
Page 5
Math 99 – Test 1 Review Spring 2013
12.
Determine which of the following points, A(2,1), B(1, – 3) and C(– 1 , 3) if any, satisfy both pairs of
equations.
y
3x – 2y
=
=
4x – 7
4
Solution: Test each point separately A(2,1) this means x = 2 and y = 1
In equation
y
1
1
=
=
=
4x – 7
4(2) – 7
1
3x – 2y
3(2) – 2(1)
4
(works)
=
=
=
4
4
4
(works)
=
=
=
4
4
4
(Fails)
Answer = A(2,1) is on both lines and so satisfies both equations.
Test each point separately B(1, – 3) this means x = 1 and y = – 3
In equation
y
=
–3 =
–3 =
4x – 7
4(1) – 7
–3
3x – 2y
3(1) – 2(– 3)
9
(works)
Answer = B(1, – 3) is not on both lines and so does not satisfy both equations.
Test each point separately C(– 1,3) this means x = – 1 and y = 3
In equation
y
3
3
=
=
=
4x – 7
4(–1) – 7
– 11
(Fails)
Answer = C(– 1,3) is not on the first line and so does not satisfy both equations.
13. Graph the line 3x + 2y = 6 by plotting at least 3 points.
y
x
y
0
3
2
0
4
–3
x
14. What is the slope and y-intercept of the line with equation 4x + 2y = 20
4x + 2y
2y
y
=
=
=
20
– 4x + 20
– 2x + 10
Slope = m = – 2
y -intercept is (0,10)
Page 6
Math 99 – Test 1 Review Spring 2013
15. (a)
On the grid opposite graph the following
lines
2x + 3y
4x – 2y
=
=
y
12
8
Working :
Table of values :
x=0
x=3
x=6
0
For 4x – 2y = 8 choose
For 2x + 3y choose
y=4
y=2
y=0
y=–4
y=–2
y=0
x=0
x=1
x=2
15.(b)
From the graph above what is the solution to this system of equations.
Solution is the point (3,2)
16.(a)
Solve the system of equations
y
Substitute y = 3x – 1 into the equation
y
=
=
3x – 1
5x – 5
y
=
3x – 1
3x
– 2x
x
Substitute x = 2 into y = 3x – 1 = 3(2) – 1 = 6 – 1 = 5
16.(b)
Solve the system of equations
Rearrange the equation 2x + y
Substitute y
=
=
=
=
=
by using the substitution method.
5x – 5
=
5x – 5
=
5x – 4
=
–4
=
2
add 1 to both sides
subtract 5x from both sides
divide both sides by – 2
So the Solution is the point (2,5)
2x + y
= 1
by using the substitution method.
3x + 4y = – 1
1 so that it is in the form y = ....... so 2x + y = 1 becomes y = 1 – 2x
= 1 – 2x into the equation
Substitute x = 1 into 2x + y
2(1) + y
2+y
y
x
1
1
1
–1
3x + 4y
3x + 4(1 – 2x)
3x + 4 – 8x
– 5x + 4
– 5x
x
= –1
= –1
=
–1
=
–1
=
–5
=
1
So the Solution is the point (1,– 1 )
Page 7
Math 99 – Test 1 Review Spring 2013
16.(c)
Solve the system of equations
Re arrange 5
=
y – 3x
Substitute y
=
3x + 5 into
to get
5x – 2y
5 =
=
–7
y – 3x
5
y – 3x
y
=
=
=
by using the substitution method.
y – 3x
5
3x + 5
5x – 2y
5x – 2(3x + 5)
5x – 6x – 10
– x – 10
–x
x
=
=
=
=
=
=
–7
–7
–7
–7
3
–3
Use x = – 3 in equation y = 3x + 5 = 3(– 3) + 5 = – 9 + 5 = – 4
The solution to the above system of equations is
16.(d)
Solve the system of equations
(– 3, – 4)
4x – 2y
y
Substitute y = 2x – 1 into the first equation
=
=
–7
2x – 1
4x – 2y
4x – 2(2x – 1)
4x – 4x +1
1
=
=
=
=
by using the substitution method.
–7
–7
–7
–7
Since 1 = – 7 is wrong we can conclude that this system of equations will have no solutions.
This would happen when the two lines are Parallel.
17.(a)
Solve the system of equations
Solution:
5x – 2y =
2x + 4y =
8
8
5x – 2y
2x + 4y
=
=
8
8
equation1….. multiply by 2
equation 2 ….leave alone
Substitute x = 2 into equation
2x + 4y
2(2) + 4y
4 + 4y
4y
y
=
=
=
=
=
by using the addition method.
10x – 4y
2x + 4y
12x
x
=
=
=
=
16
8
24
2
add equations
8
8
8
4
1
So the Solution is the point (2,1)
Page 8
Math 99 – Test 1 Review Spring 2013
17.(b)
Solve the system of equations
2x – 2y
4x + 3y
=
=
10
1
2x – 2y
4x + 3y =
multiply by – 2
put y = – 3 into equation
=
10
–1
– 4x + 4y
4x + 3y
7y
y
4x + 3y
4x + 3( – 3)
4x – 9
4x
x
=
=
=
=
=
–1
–1
–1
8
2
=
=
–1
y–2
by using the addition method.
=
=
=
=
– 20
–1
– 21
–3
add the equations
So the Solution is the point (2, – 3 )
17.(c)
Solve the system of equations
4x
4x
=
5x – 2y
4x – y
=
–1
y–2
=
=
– 10
–2
Put x = 2 into equation
by using the addition method.
multiply both sides by 10
subtract y from both sides
5x – 2y
4x – y
=
=
– 10
–2
multiply both sides by – 2
Add the equations
5x – 2y
– 8x + 2y
– 3x
x
=
=
=
=
– 10
4
–6
2
4x
4(2)
8
10
=
=
=
=
y–2
y–2
y–2
y
Solution is (2,10)
18.
A truck rental agency charges a daily fee plus a mileage fee. Julie was charged $85 for two days and 100
miles and Christina was charged $165 for 3 days and 400 miles. What is the agency’s daily fee and what
is the mileage fee?
Julie was charged $85 for two days and 100 miles
Christina was charged $165 for 3 days and 400 miles
2d + 100m
3d + 400m
=
=
85
165
Put m = 0.15 into the equation
multiply by – 3
multiply by 2
Add the equations
2d + 100m
3d + 400m
=
=
85
165
– 6d – 300m
6d + 800m
500m
m
=
=
=
=
– 255
330
75
0.15
2d + 100m
=
2d + 15
2d
d
85
=
=
=
85
70
35
So solution is d = $35 per day and m = $0.15 per mile
Page 9
Math 99 – Test 1 Review Spring 2013
19. The plumber charges a fixed rate for turning up at your house plus a charge per
hour for the work done From previous jobs that the plumber has done it was found that a 4 hour job
will cost a total of $283 while a 6 hour job will cost a total of $387.
What is the fixed rate and the charge per hour?
Let x
y
=
=
fixed amount
charge per hour
From the information a 4 hour job will cost a total of $283 we get the equation x + 4y
From the information a 6 hour job will cost a total of $387 we get the equation x + 6y
Use the substitution method so we rearrange
to get
Use x = 283 – 4y into equation
x + 4y
x
x + 6y
283 – 4y + 6y
283 + 2y
2y
y
=
=
=
=
=
=
=
=
=
283
387
283
283 – 4y
387
387
387
104
52
Use y = 52 in equation x = 283 – 4y = 283 – 4(52) = 283 – 208 = 75
So the plumber had a fixed amount of $75 plus he charged $52 per hour.
20. By weight one alloy of brass is 70% Copper and 30% Zinc. Another Alloy is 40% Copper and 60% Zinc. How
many grams of each alloy would need to be melted and combined to obtain 600 grams of a brass alloy that is
60% Copper and 40% Zinc?
x
y
=
=
amount of alloy one
amount of alloy two
x+y
0.7x + 0.4y
=
=
(70% Copper and 30% Zinc)
(40% Copper and 60% Zinc)
600
0.6(600) =
Rearrange equation 1:
0.7x + 0.4 (– x + 600)
0.7x – 0.4x + 240
0.3x + 240
0.3x
x
and y
360
to get
=
=
=
=
=
=
Equation 1:
Equation 2:
y = – x + 600 and substitute this into equation 2:
360
360
360
120
400 grams
200 grams
So we mix 400 grams of alloy one with 200 grams of alloy 2.
Page 10