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```							           January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

The following questions open response. Please work out the best answer and circle it. You MUST
SHOW ALL STEPS AND WORK OR YOUR ANSWER WILL BE WRONG!! Give examples
where necessary. Show justifications where necessary. Write all formulas used.

Bonus Questions: Write the correct formula on the line next to the phrase: ( 1 point each )
2
Standard Form ______________________ Slope-intercept Form _____________________
Point-slope Form _____________________ Sum of Squares Pattern ______________________
Difference of Squares Pattern ____________________ Quadratic Equation __________________________
Quadratic Formula ______________________________ Axis of Symmetry _________________________
Discriminant ________________________

1.     Which property is shown by 8(4  5)  8  4  8  5        ?   (Standard 25.2)
This problem distributes the 8 to what is in parenthesis, so this is the Distributive Property.

2.     When is the following statement true?         (Standard 1.1)

The opposite of a number is more            Try some examples. Use a positive number, a negative
than the original number.                   number, and 0. The opposite of 8 is -8, which is less, but
the opposite of -4 is 4, which is more. The opposite of 0 is
0, which is the same. This is true for negative numbers
only. x < 0

3.     Evaluate the expression - |-9|. (Standard 2.0)

First, find out what the value of (-|-9|) is. |-9| = 9, so the value is -9

4.     What is the reciprocal of 3 ?       (Standard 2.0)
First, make a fraction out of a whole number by adding a 1 underneath as a denominator, which
3                                                                  3          1
is     . The reciprocal of a fraction turns the fraction upside down.  becomes - .
1                                                                  1          3

5.     What is the 6 minus the opposite of -8? (Standard 2.0)

Write the expression first. 6 - -( -8). Remember that – (-8) changes to +8. The expression
becomes 6 - 8 = -2
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

6.    Graph the inequality | x  3 | 1 (Standard 3.0)

-6   -5   -4   -3    -2      -1   0

Absolute value inequalities always have 2 answers. You have to rewrite the problem 2 ways
to solve. Notice the second inequality turns the inequality sign around and changes the
number to the opposite sign. Solve for x. Mark a closed circle if the inequality has an
equal sign it, an open circle if there is no equal sign in it.
1st inequality             2nd inequality              x < -2 puts an open circle at -2
|x + 3| < 1     Look!     |x + 3| > -1                 x > -4 puts an open circle at -4
x+3<1                       x + 3 > -1
-3 -3                       -3 -3
x < -2        OR           x > -4

1
3
7.    *What is 125 ?
1
125 means the same as the cube root of 125, or 3 125 . The cube root means “What number
3

multiplied by itself 3 times equals 125?” 1*1*1 = 1, 2*2*2 = 8, 3*3*3 = 27, 4*4*4 = 64,
5*5*5 = 125. The answer is 5.

2 3
8.    Simplify (3 ) .

When raising a power to another power, you multiply the exponents.
(32 )3  323  36  3  3  3  3  3  3  9  9  9  81  9  729

9.    a.    True or False? The order in which you add 2 numbers does not affect the answer.

b.    True or False? The order in which you subtract 2 numbers does not affect the answer.

c.    True or False? The order in which you multiply 2 numbers does not affect the answer.

d.    True or False? The order in which you divide 2 numbers does not affect the answer.
Put some numbers in and see what happens.
6336               6-3  3-6         6•3  3• 6         6  3=3  6
1
9  9 True           3  -3 False 18  18 True           3  False
2
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

10.     Simplify 4  3  x   3  4 x  7  (Standard 4.0)

4(3  x)  3(4 x  7) Distribute
4 • 3 - 4 • x  3 • 4 x - 3 • 7 Simplify
12 - 4 x  12 x - 21 Gather like terms
8x - 9

11.    Solve the following equation. (Standard 4.0)
5 x   6 x  9    2 x  8

5 x   6 x  9    2 x  8  Distribute the negative sign
5 x  6 x  9 - 2 x - 8 Simplify
5 x  4 x  1 Get rid of smallest variable Subtract 4x
-4 x - 4 x
x 1

12.   Solve the following equation. (Standard 4.0)
1
 6 x  9   12
3

Remember to multiply the whole equation by the denominator 3 before you start.
1
(6 x  9)  12 Get rid of the fractions by multiplying the equation by 3
3
1               
3   (6 x  9)  12  Multiply by 3
3               
1
(3  )(6 x  9)  36 Simplify
3
6 x - 9  36 Add 9 to both sides
+9 +9
6 x  45 Divide both sides by 6
6x 45
=        Simplify
6       6
45               3
x         Divide by
6               3
15
x
2
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

13.   Solve for x.    9  12  2  2x  Standard 4.0)

Distribute the 12 to the parenthesis.
9  12(2  x) Distribute the 12 to everything inside the parenthesis.
9  12  2  12  x Simplify using PEMDAS
9  24 - 24 x Subtract 24
-24 -24
-15  -24 x Divide by -24 (Turn sign around negative division)
-15
 x Turn the inequality and the sign around (x on left)
-24
15
x
24

14.   Simplify.      – 7(-2x +4) –2x +1     (Standard 4.0)

Remember to distribute first because of PEMDAS.
7(2 x  4)  2 x  1 Distribute the -7 to everything inside the parenthesis.
7  (2) x  7  4  2 x  1 Simplify
14 x - 28 - 2 x  1 Combine Like Terms
14 x - 2 x - 28  1

15.   Solve the inequality 4x + 10  6x - 8.       (Standard 5.0)
Solve for x. Get x by itself.
4 x  10  6 x  8 Get rid of the smallest variable.
-4x       -4x
10  2 x  8 Add 8
+8       +8
18  2 x Divide by 2
18
 x Change sides to move x to the left and turn the inequality sign around.
2
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

1
16.   16. Solve for n.        n  7  12 (Standard 5.0)
5
Get rid of adding or subtracting first.
1
n  7  12 Subtract 7
5
-7 -7
1
n  5 Get rid of dividing by 5 by multiplying by 5
5
1
5  n  5  5 Cancel the 5's on the left side
5
n  25

17.   Solve for b.   9b  5 = 27  7b     (Standard 5.0)
Remember to get rid of the smaller variable first by doing the opposite.

9b  5  27  7b Get rid of the smallest variable. Add 7b
7b           7b
16b  5  27 Add 5
5 5
16b  32 Divide by 16
b2

18.   Solve for x: 5.9  3.5x  2.8x (Standard 5.0)
If you don’t like decimals, multiply everything by 10 to remove them before you start.
5.9  3.5 x  2.8 x Multiply by 10 to remove the decimal points
10(5.9  3.5 x  2.8 x) Distribute
59  35 x  28 x Get rid of the smaller variable. Subtract 28x
- 28 x - 28 x
59  7 x  0 Subtract 59
-59         - 59
7 x  59 Divide by 7
59
7
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

19.    Solve the inequality     13  4x – 7 < 17     (Standard 5.0)
Solve this like a regular inequality, but you must do the same thing to all 3 parts.

13  4 x  7  17      Add 7 to each part
+7       +7 +7
20  4 x  24 Divide each part by 4
20 4 x 24
      Simplify
4   4   4
5 x6

20.    A vase has 15 flowers in it now. Riposo adds 9 flowers to the vase every day until it holds 51 flowers.
If d represents the number of days Riposo added flowers to the vase, write the equation for this
problem, then solve it for the number of days. (Standard 5.0)
This problem makes a linear equation. You have a constant number, 15, the number of
flowers already in the vase. You have a variable, 9, because 9 flowers get added every
day, and it depends on the number of days. You have an answer given, 51. Write the
equation:
15  (9  the number of days)  51
15  9d  51 Subtract 15
15        -15 Simplify
9d  36 Divide by 9

21. DeJown has six less than four times the number of books that Sally has. If Sally has 11 books, how
many books does DeJown have? Write the equation first. (Standard 5.0)

You need to use some thinking skills to find out what information they give you. Sally has 11
books so S = 11. Eric has six less than four times the number Sally has, so E = 4S-6. Substitute
11 for S in the second equation,
E  4S  6 Substitute
E  4(11)  6
E  44  6
E  38 books
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

22. An auto repair shop charges \$1,737.50 for a new engine, including labor. 7.5 hours of labor are
required to install the engine. If the shop charge \$85 an hour for labor, what is the cost of the engine without
labor included?

The total cost with labor is \$1737.50. Subtract the cost of the labor (7.5 hours times \$85) and
you’ll get the cost of the engine.
e  1737.50  (7.5  85) Write the equation e  engine cost
e  1737.50  637.50
e  \$1100

23.    A cheetah has been clocked at a rate of 71.3 miles per hour. If it continues to run at this rate for 1.5
hours, how far would it run? (Standard 15.0)
Multiply 71.3 times 1.5 = 106.95. Since this is distance, it is in miles, NOT square miles.
Square miles is area, miles is distance.

24.    A donut machine takes .75 hours to heat up. Once it begins to make donuts, a donut machine makes
300 donuts an hour. How many hours does it take, starting with a cold oven, to make 225 donuts?
(Standard 15.0)

Find how long it takes to make 225 donuts at 3000 donuts per hour first, then add the warm-up
225
time. .        .75 hours
300
The total time is .75 hours + .75 hours warm-up, or 1.5 hours.

25.    Jesikaa earned \$444 for working 18 hours. How much did she earn per hour?          (Standard 15.0)

Divide 18 into 444, which equals \$24.67 dollars per hour.

26.    What is the y-intercept of the following equation? 6x + 3y = 18        (Standard 6.0)

To find the y-intercept, you need to put this equation in slope-intercept form by solving for y.
6 x  3 y  18 Subtract 6x
-6 x        - 6x
3 y  -6 x  18 Divide by 3
6       18
y  - x            Simplify
3        3
y  - 2 x  6 The intercept is 6
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

27.   What inequality does the graph below represent?
y                                First find the y-intercept
which is 3, and graph this on
4
the y-axis.
Next, find the slope. Find
2                                     another point on the line that
meets where an x-line and y-
-5                                         line cross. You can use the
0                             x
point (1, -2 ). Draw the
-6     -4      -2             0       2   4     6
triangle to find the x-and-y-
values. The slope is
-2
y  value        -5 down
or          or
1                                     x  value         1 right
-4
5 . Since the line is a
solid line, it uses  or  .
Next, since the inequality
shaded over the line, use  .
The inequality becomes
y  5 x  3 .

28.    The intercepts of a line are (0,5) and (-4,0). Draw the graph of this line. (Standard 6.0)
Then find the equation of the line.
y
First find the slope by
4
making a triangle with the
4                                                  line. The slope is the
5                                                                  rise y  value . In this
or
2                                        run x  value
problem, the y-value is 5
0                                x      and the x-value is 4, so the
-6      -4         -2         0       2   4      6                           5
slope is     .
4
-2
Next find the y-intercept, or
where the line crosses the y-
-4                                       axis, which is at 5.
Write the equation as
y  mx  b , where m is
the slope and b is the y-
5
y     x5
4
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

29.   What is the equation of the line below? (Standard 6.0)
First find the slope by
1y                                  making a triangle with the
line. The slope is the
4
rise y  value . In this
or
run x  value
4
2                                      problem, the y-value is 3
and the x-value is 1, so the
x
4
0                                      slope is     .
-6     -4      -2         0       2    4         6                      1
Next find the y-intercept, or
-2                                      where the line crosses the y-
axis, which is at 1.
Write the equation as
-4
y  mx  b , where m is
the slope and b is the y-
y  4x  1

30.    Graph the following inequality. y < –3x + 4 (Standard 6.0)
First find the y-intercept,
y                                  which is the number without
an x, or 4, and graph this on
4
the y-axis.
-3                                           Next, find the slope, which is
2
the number in front of x. If
this is NOT a fraction yet, put
a 1 under the number and
0                              x       make it a fraction. The slope
-6     -4      -2         0   1   2   4          6
3
is -4, and is       The
-2                                                       1
y  value             3 down
means to go
x  value             1 right
-4
from the y-intercept, and put
a second point. Then draw a
line with a ruler going
through both points. Make it
a dashed line, because < or >
use a dashed line, while
 or  use a solid line.
Next, since the inequality is <
line.
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

31.   What is the inequality of the graph below? (Standard 6.0)
First find the y-intercept
which is -2, and graph this
y
on the y-axis.
4
Next, find the slope. Find
another point on the line that
meets where an x-line and y-
2                                         line cross. You can use the
point (-4, 1 ). Draw the
x
triangle to find the x-and-y-
0
-6         -4       -2              0   2    4      6
values. The slope is
-3                                                                 y  value     -3 down
or            or
-2
x  value      4 right
3
4              -4
. Then draw a line with
4
a ruler going through both
points. Make it a dashed
line, because < or > use a
dashed line, while  or 
use a solid line. Next, since
the inequality is < (less than),
The inequality is
3
y      x2
4

32.    Find 3 points that lie on the line defined by 4 x  8 y  2 .   (Standard 7.0)

The easiest way is to make a T-Table of x-and-y-values, then put in an x-value, and find the y-value.
First, substitute 1 for x in the equation, then 2, then 3..
4 x  8 y  2 Substitute               4 x  8 y  2 Substitute          4x+8y=2 Substitute
4 •1  8 y  3 Simplify                4 • 2  8 y  2 Simplify          4  3+8y=2 Simplify
4  8 y  2 Subtract 4                   8  8 y  2 Subtract 8           12+8y=2 Subtract 12
-4        - 4 Simplify                   -8       - 8 Simplify             -12     -12 Simplify
8 y  -2 Divide by 8                         8 y  -6 Divide by 8              8y=-10 Divide by 10
-2 1                                        -6 3                              -10 5
y                                           y                                y      
8   4                                        8   4                               8   4
-1                                       -3                                        -5
1st point (1,         )                     2nd poin t (2,      )               3rd poin t (3,            )
4                                        4                                          4
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

33.    Which equation is represented by the graph?    (Standard 7.0)

First find the slope by making a triangle with the
line. The slope is the
rise y  value . In this problem, the y-value is 4
or
run x  value
4 2
and the x-value is 6, so the slope is     .
6 3
Next find the y-intercept, or where the line crosses
the y-axis, which is at -4.
Write the equation as y  mx  b , where m is the
slope and b is the y-intercept. The answer is
2
y  x4
3

34.   In the graph below, what is the equation of the line? (Standard 7)
First find the slope by
y                                       making a triangle with the
1                                                line. The slope is the
6
rise y  value . In this
or
run x  value
4            4                                            problem, the y-value is 4
and the x-value is 1, so the
2
4
slope is     .
1
x
Next find the y-intercept, or
0
-6      -4       -2            0   2    4      6
where the line crosses the y-
axis, which is at 6.
-2                                           Write the equation as
y  mx  b , where m is
the slope and b is the y-
y  4x  6
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

35.       Find the equation of the line passing through the points (1, 10) and (-1, 4) in slope-intercept form and
standard form. (Standard 7.0)

This type of problem has several steps, but we’re still looking for the y-intercept and the slope.
x1 y1      x2 y2
First: Find the slope. Write the points in a line (1, 10) and (-1, 4). In these points, the first x = 1 and the
first y = 10. The second x = -1 and the second y = 4. Write this as x1 = 1, y1 = 10, x2 = -1, and y2 = 4.
y  y 4  10 6
Second: Put the numbers you found into the slope equation m  2 1                         3 . The slope is 3.
x2  x1 1  1 2
Third: Now that you know the slope, substitute one of the points into the point-slope from to find the

y  y1  m( x  x1 ) Point-slope form. Substitute
y -10  3( x -1) Simplify & distribute
y  10  3 x - 3 Add 10
+10       +10 Simplify
y  3 x  7 Slope-intercept form.
Fourth: Change slope-intercept form into standard form. Standard form has the x-and-y-values on the
same side.

y  3x  7 Slope-intercept form.
3x - 3x Simplify
-3x  y  7 Standard Form

36.       Does the point (4, -3) belong on the line of 3x – y = 15?        (Standard 7.0)

Substitute 4 for x and -3 for y into the equation to see if it is true. If it is true, the point is a solution.
3 x  y  15 Substitute the values
(3 • 4) - (-3)  15 Simplify
12  3  15 Simplify
15  15 Solution
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

37.    What is the equation of the line with slope -4 that passes through point (6, 5) in point-slope form and
slope-intercept form? (Standard 7.0)
If you already know a point and a slope to an equation, use the point-slope form to find the equation.
The point-slope form is ( y  y1 )  m( x  x1 ) .
x1 y1
First: Find x1 and y1. For the point (6, 5) x1 = 6 and y1 = 5. m is the slope, which is -4.

Substitute both of these into the point-slope form ( y  y1 )  m( x  x1 ) .

( y  y1 )  m( x  x1 ) Substitute
( y  5)  4( x  6) Answer in point-slope form.
( y  5)  4( x  6) Distribute -4
y - 5  -4 x  24 Add 5
5           5 Simplify
y  -4 x  29 Answer in slope-intercept form

38.    Two lines with the same slopes and different y-intercepts are ________________ lines.
(Standard 8.0)
Lines with the same slopes but different y-intercepts are parallel
lines
Two lines with reciprocal and opposite slopes are _____________________ lines.
Lines with reciprocal and opposite sign slopes are perpendicular
Two lines with equivalent slopes and the same y-intercept are __________________ lines.
Lines with equivalent slopes and the same y-intercept are identical lines.

3
39.    a.     Write an equation of a line parallel to y   x  6 . (Standard 8.0)
5

3
b.     Write an equation of a line perpendicular to y   x  6 . (Standard 8.0)
5

Parallel: You need to write a linear equation with the same slope but a different y-intercept.
3
y      x7
5

Perpendicular: You need to write a linear equation with the opposite sign reciprocal slope with
any y-intercept.
5
y  x7
3
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

40.      Graph the equations and find the solution of the system of equations shown below. (Standard 9.0)

2x  y  6                                                       First, change both equations to
slope-intercept form.
x y 4
y
2 x  y  6 Subtract 2x
6                                   -2 x       - 2x
y  -2 x  6 1st equation
4
x y4        Subtract x
2                                   -x      -x
y  -1x  4    2nd equation
0                             x
-6      -4     -2         0    2      4       6
Next, graph both equations.
-2

Solution is                           y  2 x  6 has a y-intercept of 6
(2, 2)
-4                                  and a slope of -2.
-1
y  -1x  4 has a slope of       and
1
a y-intercept of 4. The solution is
where the 2 lines cross, at the point
(2 , 2)

41.        What is the solution of the system of equations shown below? (Standard 9.0)
x y 6
x  3y  2

x y6                                                       x  y  6 Substitute
+ -1( x  3 y  2) Multiply by -1 to get opposite x-values     +   x  1  6 Subtract 1
1x  3 y  2                                                     -1 -1
+
x y6                                                           x5
4y  4 Add the variables               Write the solution as a point
4y 4
Add the x’s            Divide by 4                   where x  5 and y  1 (5,1)
to get 0       4 4
y 1
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

42.   Graph the system of inequalities below and identify the solution. (Standard 9.0)
First find the y-intercept of
y  3x                                                               y < 3x. Since there is no y-
y  1x  6                                                          intercept, put in 0. The
inequality becomes
y < 3x + 0. The y-intercept is
y                                       0, so put a point on 0. The
8
slope is 3  3 up , so from
y < -1x + 6                                y < 3x                            1 1 right
0, go up 3 and right 1 and
6
make another point. The
inequality sign is <, so this is
4
a dotted line, and < means
less than, so shade below the
2                                            line.

0                                 x          First find the y-intercept of
-6        -4       -2        0      2         4      6
y < -1x + 6. The y-intercept
-2                                           is 6, so put a point on 6. The
Solution
-1 -1 down
slope is              , so
1   1 right
from 6, go down 1 and right
1 and make another point.
The inequality sign is <, so
this is a dotted line, and <
below the line.

43.   |x|= - 11 When is this statement true? (Standard 25.3)

Try some numbers.
|11| 11
|
| 11| 11
There is no answer because absolute value signs remove the positive or negative sign, so there
can never be a negative answer.
No solution.

44. *Two birds flew from their nest in opposite directions. One bird averages 10 miles per hour and the
other bird averages 15 miles per hour. How many hours will it take for the birds to be 87.5 miles apart?

If 2 birds fly in opposite directions, they are getting farther apart. Add the speeds to find out
how far apart they are after 1 hour. Then divide 87.5 miles by the distance apart after 1 hour, or
87.5
25.          3.5 hours
25                                  10      +   15 = 25
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

45. McWilly can eat a super-large pizza by himself in 30 minutes. McDilly can eat the same pizza by
himself in 40 minutes. If they both eat the same pizza at the same time, how long will it take to eat the
whole thing?

46.    Macho rode his motorcycle for 3 hours at an average speed of 50 miles per hour. For the first 2
hours, his average speed was 45 miles per hour. What was his average speed for the last hour of his
trip?

First, find out how far the entire trip was. If he traveled for 3 hours at 50 miles per hour, the
entire trip went 150 miles.
Second, find out how far he has gone already. If he went 2 hours at 45 miles per hour, he
already went 90 miles. Subtract 90 from 150 to find out how far her needs to go the last hour.
150 – 90 = 60. The last hour he needs to travel 60 miles per hour.

47.    BobbieJean pulls 30 coins out of her purse. Each coin is either a dime or a quarter. If she has a total
of \$5.85, how many dimes does he have? How many quarters? Solve this as a linear
system.(Standard 9.0)
You can solve this several ways. First
D             Q              30             5.85
is to just guess numbers, but that is time
5            25                            6.75
consuming. Second is to make a T-
10            20             30             6.00
Table of values. Third is to solve it as
a linear system. The easiest is to make            12            18             30             5.70
a T-Table.                                         14            16             30             5.40
There are 11 dimes and 19 quarters.                11            19             30             5.85

48.   Solve.   | 2 x  4 | 8 (Standard 3.0)

Absolute value equations usually have 2 answers. You have to rewrite the problem 2 ways to solve.
Notice the second equation changes the number to the opposite sign. Solve for x.
1st equation                      2nd equation

|2x - 4| = 8       Look!         |2x - 4| = -8
2x - 4 = 8                        2x - 4 = -8
+4 +4                              +4 +4
2x = 12                          2x = -4
x=6              OR            x = -2
January 08 – January 2010 Algebra I Practice Final Version 3 Solutions

49. Find the solutions to the following system. (Standard 9.0)
2 x  3 y  0 
              
4 x  6 y  4 

- 2 (2 x  3 y  0) Multiply by - 2 to get opposite x - values       2 x  3 y  0 Substitute
1
     4x  6 y  4                                                 2 x  (3 • )  0 Simplify
3
+  4 x  6 y  0 Add the var iables                                         2 x -1  0
12 y  4    Divide by 12                                             1  1 Add 1
12 y 4 1
                                                              2 x  1 Divide by 2
1                                                                   1
x’s to get 0      y                                                                  x
3                                                                   2
Substitute int o the first equation                           Write the solution as a po int
1          1       1 1
where x  and y  , or ( , )
2          3       2 3

50. Bethany bought a new Honda Civic Hybrid car that is supposed to get at least 50 miles per gallon. To
test this, she drove to Las Vegas and back, traveling the whole 276 miles on 5.3 gallons of gas. How many
miles per gallon did she average? (Round your answer to the nearest tenth) (Standard 15.0)

Miles per gallon means miles divided by gallons,
miles 276          miles         miles
or               52.07         52.1        . Round to 52.1                              miles
gallons 5.3         gallon        gallon                                             gallons

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