Please let me know if you find any mathematical errors on this page.  For students in my math 73 class, I will award extra credit for any mistake(s) you bring to my attention.  (To receive credit, you'll have to explain the mistake and how to correct it.)

1. (one point each) Determine whether the following statements are always, sometimes or never true.

1. If aⁿ = bⁿ, then a=b.   SOMETIMES TRUE.  (Counterexample: a=1, b=-1, n=2)
2. If x<0, then sqrt(x²) = -x.  ALWAYS TRUE.

(Note that if x<0, then -x>0, as required.  Try a few simple examples: x=-1, x=-2, etc.)
3. The polynomial b³+1 is not factorable.  NEVER TRUE.

(b³+1 factors as (b+1)(b²-b+1).)
4. 6x+5 is an equation. NEVER TRUE.

(This should be obvious -- since there's no equal sign, there's no equation.)
5. x-3 = x-3 has no solution.  NEVER TRUE.

(In fact, every real number is a solution!)
6. If a>b, then -a<-b.  ALWAYS TRUE.

(When multiplying both sides of an inequality by -1, its direction changes.)
7. (2+3)^-1 = 1/2 + 1/3.  NEVER TRUE.

(This one should require no explanation.)
8. When multiplying two powers with the same base, the rule is to add the exponents.  ALWAYS TRUE.

(This is a fundamental property of exponents which you should know.)
9. An inconsistent system of equations has no solution.  ALWAYS TRUE.

(This is the definition of an "inconsistent system.")
10. A system of two equations with the same slope and different y-intercepts has an infinite number of solutions.  NEVER TRUE.

(The situation described here is that of two parallel lines with different y-intercepts.  Two such lines will never intersect, and so the system will have no solution.)
11. Perpendicular lines have the same y-intercept.  SOMETIMES TRUE.

(Perpendicular lines may have the same y-intercept, but they certainly don't have to.)
12. f(a) represents a value in the range of a function when a is a value in the domain of the function.  ALWAYS TRUE.

(This is essentially the definition of "domain" and "range.")
13. If x^-1 is negative, then -x is positive.  ALWAYS TRUE.

(x^-1 has the same sign as x, so if x^-1 is negative, then so is x.  Thus, -x is the opposite of a negative number and is therefore positive.)
14. An equation in two variables has an infinite number of solutions.  SOMETIMES TRUE.

(This one is tricky.  It's certainly true for linear equations (y=mx+b) for example, so right away we know the answer is either "sometimes" or "always."  However, it's also possible for such an equation to have no solution.  For example, we could come up with an equation in which one side must be negative and the other positive, like -|x| - 1 = |y| + 1.)
15. If a and b are real numbers, then |a-b| = |b-a|.  ALWAYS TRUE.

(Since a-b is always the opposite of b-a, the absolute value of each of these two numbers will be the same.)
16. If a<b, a != 0, and b != 0, then 1/a < 1/b.  SOMETIMES TRUE.

(The statement true if a is negative and b is positive.  However, if a and b are both positive, then the statement is false.  To see this, try examples such as a=-1, b=1 and a=1, b=2.)
17. If a car travels 2 hours at a rate of 40 mph, and then another hour at 60 mph, then its average speed is 50 mph.  NEVER TRUE.

(Under these conditions, the car travels 140 miles in 3 hours, so its average speed is 140/3 = 46+2/3 mph.)
18. The expression 2(x+4) can be described by the statement "the sum of twice x and 4."  NEVER TRUE.

(The given statement would be written as 2x + 4.  The correct statement for 2(x+4) would be "twice the sum of x and 4.")
19. (x² + 2x + 1)/(x^1/2 -2x + 4) is a rational expression.  NEVER TRUE.

(The denominator is not a polynomial, so this is not a rational expression.)
20. Zero is a solution of 5x=2x.  ALWAYS TRUE.

 

i)

9[

(5/6) - 2 (3/8)

]

7 6

Solution: The fraction inside the square brackets simplifies to -28/9 (after multiplying the top and bottom by 24).  Multiplying by 9 gives us -28; finally, dividing by 7/6 (i.e., multiplying by 6/7) gives us our solution: -24.
 

ii)

x1/2 x1/2 - y1/2

=

x + (xy)1/2 x-y

To rationalize the denominator, multiply the top and bottom by x^1/2 + y^1/2.  Thus, the numerator is x + (xy)^1/2, and the denominator is x-y.
 

iii)	6 - x - x2 3x2 - 10x + 8	=	(2-x)(3+x) (x-2)(3x-4)
		=	-(3+x) 3x-4

(Note that (2-x) is the opposite of x-2.)
 

iv)	2x  - 6 6x2 - 15x		4x2 - 12x 18x3 - 45x2		(First, factor everything; then rewrite as multiplication.)
=	2(x-3) 3x(2x-5)	.	9x2(2x-5) 4x(x-3)		(Now cancel out all common factors of the numerator and the denominator.)
=	3/2	(Since the factors (x-3), (2x-5) and 6x2 cancel.)


 

v)	(16x2y4)1/2 (8a6b3)1/3	.	28a2b4 2xy4		(First, distribute the exponents on the left-hand side.)
=	4xy2 2a2b	.	28a2b4 2xy4		(Now cancel out all common factors, as before.)
=	28b3 y2

(Note: for the rest of #2, only solutions will be given; the steps will be omitted.  Talk to me or send me email if you have any questions about these problems.)
 

vi)

a - 3 a2 - 5a

+

a - 9 a2 - 25

=

2a+3 a(a+5)

vii)   3(32x²)^1/2 - 2x(2^1/2) + (128x²)^1/2 = 18x(2^1/2)
 

viii)

(x-1 + y-1)-1

=

xy x+y

 

Solution: P(-2+h) - P(-2)	=	[4(-2+h) + 7] - [4(-2) + 7]
	=	[4h - 1] - [-1]
	=	4h

 

i)	2x - 7 3	-	5x + 4 5	=	-x - 4 30		(Now multiply both sides by the least common denominator, which in this case is 30.)
	10(2x-7)	-	6(5x+4)	=	-x - 4		(Now solve for x.)
	-10x - 94			=	-x -4
	-90			=	9x
			-10	=	x

ii)  3x - 2 > 4  or  4 - 5x < 14
The first inequality reduces to x>2, and the second to x>-2.  In interval form, then the solution set is (2,) together with (-2, ).  The union of these two intervals is simply (-2, ).

iii) 2 - |2x - 5| = -7.
Isolate the absolute value term, to get |2x - 5| = 9.  This is actually two equations: 2x - 5 = 9, and 2x - 5 = -9.  Solving each gives us two equations: x=7, x=-2.

iv) 1 - 7/(x+7) = 3/(x+3)
Multiplying both sides by the least common denominator, (x+7)(x+3), gives us the equation:
 

(x+7)(x+3) - 7(x+3)	=	3(x+7)
x2 + 10x + 21 - 7x - 21	=	3x + 21
x2 + 3x	=	3x + 21
x	=	+ 211/2

v) -2 < 2 - 3x < 7
Add 2 to all three parts, and then divide by -3, to reduce the inequality to -5/3 < x < 4/3.  The solution, in interval notation, is [-5/3, 4/3].

vi) |x + 1| = x + 1.
This equation can only be true if x+1 is positive or 0.  (A negative number would not be equal to its absolute value.)  Therefore, the problem is really to solve the inequality x + 1 > 0; the solution set is [-1, ).

vii) x(x-12) = -27
Rewrite as x² - 12x + 27 = 0.  The left hand side factors as (x-9)(x-3), and so the solutions are x=9 and x=3.

viii) x³ - x² - 25x + 25 = 0
The left hand side can be factored:

x2(x-1) - 25(x-1)	=	0
(x2-25)(x-1)	=	0
(x+5)(x-5)(x-1)	=	0

Therefore, the solutions are x=-5,x=5 and x=1

ix) (2x+3)(x-7) = 0
The solutions are x=-3/2 and x=7.

x) (x-3)^1/3 + 5 = 0

(x-3)1/3	=	-5		(Raise both sides to the 3rd power.)
x - 3	=	-125
x	=	-122

5. Translate into a variable expression and simplify: Fifteen minus one half the sum of a number and ten.

First of all, let x stand for "the number."  Then, "One half the sum of a number and ten" is ½(x+10), and so "fifteen minus one half the sum of a number and ten" will be written 15 - ½(x+10).

6. A library charges a fine for each overdue book.  The fine is 15 cents for the first day plus 7 cents for each additional day.  Let y be the fine if the book is overdue for x days.

a) Write y as a function of x.
    y = 15 + 7(x-1)  [where we define y to be the fine in cents].  This can also be written y = 7x + 8.

b) If the fine is 78 cents, how many days overdue was the book?
    Solve the equation from part (a), with y=78:
 

	78	=	7x + 8
	70	=	7x
	10	=	x

Therefore, the book was ten days overdue.

7. For the linear equation 4x - 2y = 5, find the x-intercept, y-intercept and slope.

Solve for y to find the slope and y-intercept:  y = 2x - 5/2.  This tells us the slope is 2 and the y-intercept is -5/2.  To find the x-intercept, plug in y=0 and solve for x.  The x-intercept is 5/4.  (Note: I'll get a picture of the graph posted here later if I have time.  Email me if you need help sketching it on your own.)

8. Find the equation for:  (i) a vertical line passing through (4, -3); (ii) a horizontal line passing through (-3, 4).

Solutions: (i) x = 4    (ii) y = 4

9. Check to see if the given relation defines a function.  If it is a function, find the domain and range. (-4, 2), (-2, 2), (0,0), (3,5).

Domain: {-4, -2, 0, 2}    Range: {0, 2, 5}

10. Solve the system of equations:
3x + 4y	=	2
2x + 5y	=	1

Multiply the top equation by 2 and the bottom equation by -3, and then add (to eliminate x):
 

6x + 8y	=	4
-6x - 15y	=	-3
-7y	=	1

Therefore, y = -1/7.  Plug this back into either of the original two equations, to get x = 6/7.

11. (8x³ - 9)  (2x-3)

Use long division -- the solution is 4x² + 6x + 9 + 18/(2x-3).

12. Factor: 4y² - 15y + 9

Solution: (y-3)(4y-3)

13. The sum of two numbers is 10.  Three times the larger number is three less than 8 times the smaller number.  Find the numbers.

Let x stand for the larger number, y for the smaller number.  Then we know the following: x+y=10, and 3x = 8y-3.  The easiest way to solve this system of equations is by substitution; use y=10-x (or x=10-y).  The result should be: x=7, y=3.

14. Find the equation to a line that contains the point (-1, -3) and is perpendicular to the line 3x-5y=2.

The line 3x-5y=2 has slope 3/5 (solve for y to find the slope), and so the slope of any perpendicular line will be -5/3.  Therefore, we're looking for the equation of the line with slope -5/3 which passes through (-1, -3).  Using the point-slope equation, we get (y-(-3)) = (-5/3)(x-(-1)), which simplifies to y+3=(-5/3)x - 5/3.  Solving for y (to write the equation in slope-intercept form, which you should always do) gives us the equation y = (-5/3)x - 14/3.

15. State whether f(x)=(3x² - 6x + 9)/x is a polynomial.  Solution:  NO.


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Last modified: 12/9/99
Kurt Ludwick (ludwick@math.temple.edu)