AND COMMON MISTAKES
PROF. BOYAN KOSTADINOV
NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
Contents
1. ACT Compass Practice Tests 1
2. Common Mistakes 2
3. Distributive Properties 8
4. Properties of Fractions 9
5. Properties of Exponents 11
6. Properties of Radicals 13
7. The Slope and Equation of a Line 14
1. ACT Compass Practice Tests
If you have Internet access, you can access the online resources below through the pdf file by simply clicking on the links below.You can use these resources to practice with sample ACT Compass tests online or watch video tutorials on Google Video. If you have only a printed copy of the pdf file, you can still find these Internet resources by using the provided web links.
(1) CUNY Compass Practice Tests from Hostos Community College http://www.hostos.cuny.edu/oaa/compass/prealgebra.htm (2) Kentucky Early Math Testing Program Practice Tests
https://www.mathclass.org/wqs/k.asp?state=1 (3) Google Video Tutorial on Order of Operations
http://video.google.com/googleplayer.swf?docid=-3581910795500993427 (4) Google Video Pre-Algebra Tutorial
http://video.google.com/videoplay?docid=-2898932824775207461 (5) Google Video Tutorial on Solving Equations
http://video.google.com/videoplay?docid=4755123812601857335 Date: November 2009.
2. Common Mistakes
Common Mistake 1. A surprisingly common mistake is to incorrectly copy the problem in your exam booklet. Make sure you are working on the correct problem!
Common Mistake 2. Always put parenthesis around a negative number, especially when you have to multiply it by another number, as in this case:
5 + 2 · (−5) = 5 − 10 = −5
Never drop the parenthesis around the negative number because you will forget that you have a multiplication and you will get this instead:
5 + 2 − 5 = 2
which is wrong and has nothing to do with the original problem.
Common Mistake 3. Be very careful with the order of operations. The correct order of operations is given below:
(1) First do the operations inside the Parenthesis. (2) Then take care of Exponents,
(3) Multiplication, Division, (4) Addition, Subtraction.
Consider as an example the algebraic expression 2 + 3 · (−9). There are several mistakes you can make. Firstly, if you don’t put the parenthesis around the negative number, you will get: 2 + 3 − 9 = −4, which is wrong. Secondly, you can get the order of operation wrong:
2 + 3 · (−9) = 5 · (−9) = −45 Wrong!
after first adding 2 and 3 and then multiplying the result by (−9). This is wrong, because multiplication has a priority, so one should first multiply 3 and (−9) to get −27 and only then add 2 to the result. So, the correct thing to do is the following:
It would be different if we have the following expression (2+3)·(−9). The difference is that 2 and 3 are now inside a parenthesis, so we would have todo the operation inside the parenthesis first and then multiply:
(2 + 3) · (−9) = 5 · (−9) = −45 Correct!
Common Mistake 4. A few common mistakes are related to the properties of ex-ponents. For example, note that 2 · 32 6= (2 · 3)2 because taking the exponent has a priority over multiplication. So, if one wants to calculate 2 · 32 then one
should take the exponent first 32 = 9 and then multiply the result by 2 to get 18, that
is 2 · 32 = 2 · 9 = 18. While for (2 · 3)2, we first do the multiplication inside the
parenthesis to get 6, which we then square:
(2 · 3)2 = 62 = 6 · 6 = 36 Another common mistake related to exponents is to write
32· 35 = 310 Wrong!
It’s also wrong to write
32· 35 = 97 Wrong!
instead of using the correct property
32· 35 = 32+5= 37 Correct!
Keep in mind that the general property reads
aman= am+n Correct!
Finally, if we have to take the power of a power, it is wrong to write (x2)5 = x7 Wrong!
The correct application of the power property reads (x2)5 = x10 Correct! Remember the general property has the form:
Example 1. Consider now the following example. Evaluate the expression: 12 − 23
−2 − 3 · (2 − 5)
3
One common mistake is to write the second term in this difference as 3 · (2 − 5)3 = (6 − 15)3 Wrong!
This is wrong because we are distributing 3 inside the parenthesis pretending that we have (2 − 5) and ignoring the fact that we actually have this difference raised to the power of 3. To do it correctly, we need to do the operation inside the parenthesis first and then raise the result to 3rd power:
3 · (2 − 5)3 = 3 · (−3)3 = 3 · (−27) = −3 · 27 = −81 Correct! and only then multiply the result by 3. In our case, we have
(−3)3 = (−3) · (−3) · (−3) = (−1)3· 33 = −27
Remember that negative number raised to an odd power must be negative and a negative number raised to an even power must be positive
(negative)odd= negative (negative)even = positive
For example, (−1)2 = 1, (−1)3 = −1, (−1)4 = 1, (−1)5 = −1.
Now, let’s go back to the original example. In the first term, we must take care of the exponent in the numerator first, so write 23 = 2 · 2 · 2 = 8 and at the same time
simplify the second term as we did earlier: 12 − 23 −2 − 3 · (2 − 5) 3 = 12 − 8 −2 − 3 · (−3) 3 = 4 −2 − 3 · (−27) Note here that −24 = −42 = −2 (don’t forget the minus sign):
We have here the product of two negative numbers −3 and −27, which gives us a positive number (−3) · (−27) = 3 · 27 = 81. Finally, if you are confused about the sum −2 + 81, note that this is really the same as the difference 81 − 2, we simply take 2 away from 81 to get 79.
Common Mistake 5. Remember that the fraction AB means that we divide A by B, i.e. we have A ÷ B. For that reason, we can express division by B in terms of multiplication by the reciprocal of B, which is B1, namely
A ÷ B = A · 1 B =
A B
Consider the division when A = 15b2a3 and B = 5b2, then we have
15b3
2a
÷ (5b2)
Sometimes, students attempt to use the division rule above but since they cannot quite remember it, they would write something like this
15b3 2a ÷ (5b2) = 15b 3 2a · 5b 2 1 Wrong!
This is wrong, of course, because the division is replaced by multiplication but the reciprocal of 5b2 is not taken. Instead, 5b2 is divided by 1, which does not change
anything since any number divided by 1 is the number itself, that is 5b12 = 5b2. This
way, the division is simply replaced by multiplication while nothing else changes and this is wrong. The correct thing to do is to take the reciprocal of 5b2 when replacing division by multiplication, namely:
15b3 2a ÷ (5b2) = 15b3 2a · 1 5b2 = 15b3 10ab2 = 3b 2a Correct!
Let’s recall the rule for multiplying two fractions that is used above. We multiply the nominators and the denominators of both fractions
a b · c d = a · c b · d
In the example above, we reduce the fraction 1510 = 5·35·2 = 32 by canceling out the common factor of 5. Remember that if we have the same number above and below the bar of a fraction then we can cancel this number:
a · b a · c =
b c
because the bar of the fraction represents a division. The property of exponents, used above to get the final result, is
bm
bn = b m−n
which we apply to our example to conclude that b3
b2 = b
3−2= b1 = b
Common Mistake 6. One of the most common mistakes when dealing with frac-tions is the following ‘rule’ that students invent to add unlike fracfrac-tions:
a b + c d = a + c b + d Wrong!
It is very easy to see that this ‘rule’ is not correct by checking a simple example. Take a = 2, b = 1, c = 3 and d = 1 and if this ‘rule’ is correct we should get a true statement: 2 1 + 3 1 ? = 2 + 3 1 + 1 2 + 3=? 5 2
which is clearly a false statement, since 5 6= 52, so the ‘rule’ cannot be correct. The correct rule we get by cross-multiplying numerators by denominators and the sum of the two products gives us the new numerator, while the new denominator is just the product of the two denominators:
a b + c d = a · d + c · b b · d
Of course, given specific numbers, one can also look for the LCD (least common denominator) but in general many students find the LCD concept more difficult. For example, let’s add the two fractions using the correct rule:
5 6 + 11 12 = 5 · 12 + 11 · 6 6 · 12 = 60 + 66 6 · 12 = 126 6 · 12 = 6 · 21 6 · 12 = 21 12 = 3 · 7 3 · 4 = 7 4
Alternatively, it is easy, in this case, to find the LCD, which is 12. The next step is to write the first fraction as an equivalent fraction having denominator of 12 and then we can easily add the like fractions. That’s why we multiply by 2 the numerator and denominator of the first fraction:
5 6 + 11 12 = 5 · 2 6 · 2+ 11 12 = 10 12+ 11 12 = 10 + 11 12 = 21 12 = 7 4
Remember the rule for adding like fractions (with the same denominators): a b + c b = a + c b which we use above to add the like fractions:
10 12 + 11 12 = 10 + 11 12
Common Mistake 7. Some common mistakes related to radicals are writing: √
x16 = x4 or √x9 = x3 Wrong!
We can check easily if our guess is correct by simply using the definition of square root, which in the first case would mean that if we take the square of our guess x4, we should get what is inside the radical: (x4)2 should be equal to x16. However, (x4)2 = x4·2 = x8 6= x16, so our guess x4 cannot be correct. I can only guess that the
logic that leads to the wrong claims above goes along these lines √ x16= x √ 16 = x4 or √x9 = x √ 9 = x3 Wrong!
The correct rule to apply in the case of an even power under the radical sign is: √
It is easy to see that x8 is the correct answer because if we square it, we get
(x8)2 = x8·2 = x16 and x16 is what’s inside the radical sign. In the second case, when
we don’t have an even power, 16 is an even power but 9 is not, we need to split the odd power in order to get an even power:
√
x9 =√x8· x =√x8·√x = x82 ·
√
x = x4·√x Correct! Here for a, b > 0, we used the product rule for radicals
√
a · b =√a ·√b
and we can generalize the rule we used above for any positive even power under the radical sign, like 16 or 8
√
xeven = xeven2
3. Distributive Properties
Let a, b, c, d, e be any numbers, positive (negative) or terms containing vari-ables. If we have to multiply two factors, we must multiply every term in the first factor by every term in the second factor. It is useful to draw arrows indicating all possible products. The following distributive properties are often used:
• a · (b + c) = a · b + a · c
• (a + b) · (c + d) = a · c + a · d + b · c + b · d
• (a + b) · (c + d + e) = a · c + a · d + a · e + b · c + b · d + b · e
Example 2. Multiply and combine like terms in the following examples: (1) 2x · (4x2+ 3) = 2x · 4x2+ 2x · 3 = 8x3+ 6x (2) −2 · (4 − 3) = −2 · 4 − 2 · (−3) = −8 + 6 = −2 (3) (x − y) · (x + y) = x · x + x · y − y · x − y · y = x2− y2 (4) (x − 4) · (x2+ 2x − 2) = x · x2+ x · 2x + x · (−2) − 4 · x2− 4 · 2x − 4 · (−2) = = x3+ 2x2 |{z} −2x − 4x2 |{z} −8x + 8 = x3− 2x2− 10x + 8
4. Properties of Fractions
a · b c · b =
a c
Example 3. This applies to numerical as well as algebraic fractions. Remember that a fraction represents a division, so if we have the same term (b in the formula) above and below the division line, being the bar of the fraction, then we can cancel this term, as we are really dividing the term by itself.
• 5·7 7·2 = 5 2 • (x−1)·(x+2)(x+2)·(x+5) = (x−1)(x+5) a −b = −a b = − a b
Example 4. We can move the minus sign from the denominator to the numerator or we can place it in front of the fraction. If both the denominator and the numerator are negative, we can cancel out the minus.
• 5 −7 = −5 7 = − 5 7 • −5 −7 = 5 7 a b · c d = a · c b · d
Example 5. The product of two fractions is a fraction whose numerator is the prod-uct of the two numerators (a · c above) and whose denominator is the prodprod-uct of the two denominators (b · d above).
• 5 7 · 2 3 = 10 21 • (x−4)(x+1) ·(x−2)3 = (x−4)·(x−2)3(x+1) a b ÷ c d = a b · d c
Example 6. We divide one fraction (the divident) by another (the divisor) by mul-tiplying the divident by the reciprocal of the divisor.
Remember, the reciprocal of c d is d c • 5 7 ÷ 2 3 = 5 7 · 3 2 = 15 14 • (x−3)(x+5) ÷(x−2)(x−1) = (x−3)(x+5) · (x−1)(x−2) = (x−3)·(x−1)(x+5)·(x−2) a d + b d = a + b d
Example 7. We add two like fractions (having the same denominators) by adding the numerators and keeping the common denominator.
• 5 7 + 4 7 = 9 7 • 5 x + 2 x = 7 x a d − b d = a − b d
Example 8. We subtract two like fractions (having the same denominators) by sub-tracting the corresponding numerators and keeping the common denominator.
• 5 7 − 4 7 = 1 7 • 5 x − 2 x = 3 x a b + c d = a · d + c · b b · d
Example 9. We add two unlike fractions having different denominators b 6= d, by transforming them first into equivalent like fractions, having the same denominators:
• 5 7 + 2 3 = 5·3 7·3 + 2·7 3·7 = 15+14 21 = 29 21 • 2 + 5 x = 2·x x + 5 x = 2x+5 x • 2 x2 + 5 x = 2 x2 + 5·x x·x = 2 x2 + 5·x x2 = 2+5x x2 a b − c d = a · d − c · b b · d
Example 10. We subtract two unlike fractions having different denominators b 6= d, by transforming them into equivalent like fractions, having the same denominators:
• 5 7 − 2 3 = 5·3 7·3 − 2·7 3·7 = 15−14 21 = 1 21 • 2 − 5 x = 2·x x − 5 x = 2x−5 x • 2 x2 − 5 x = 2 x2 − 5·x x·x = 2 x2 − 5·x x2 = 2−5x x2 5. Properties of Exponents xn= n z }| { x · x · · · x
Example 11. Raising a number (or an algebraic expression) to some positive integer power is the same as multiplying this number (or an algebraic expression) by itself as many times as the positive integer power.
• (−2)3 = (−2)(−2)(−2) = −8
• (x − 2)3 = (x − 2)(x − 2)(x − 2)
• (−2)4 = (−2)(−2)(−2)(−2) = 16
xmxn= xm+n
Example 12. Observe that all powers in the formula above have the same base x, which could be a number or a more general algebraic expression. To multiply two powers having the same base (x above), we add the exponents and keep the base.
• 23· 25 = 23+5= 28 → the base here is 2
• (x − 2)3· (x − 2)4 = (x − 2)3+4= (x − 2)7 → the base here is (x − 2)
• a4· a5 = a4+5 = a9 → the base here is a
• 24· 35 6= (2 · 3)4+5 → bases are different, 2 and 3
(xm)n= xm·n
• (23)5 = 23·5 = 215 • ((x − 2)3)4 = (x − 2)3·4 = (x − 2)12 • (a4)5 = a4·5 = a20 • (24)5 6= 24+5 xm xn = x m−n
Example 14. The quotient of two powers having the same base (x above) and any exponents (m and n above) is the base to the power that is the difference of the two exponents (m − n above), where the exponent that is below the bar of the fraction (n above) is subtracted from the exponent above the bar (m above).
• 25 23 = 2 5−3 = 22 = 2 · 2 = 4 → base is 2 • x5 x3 = x5−3 = x2 → base is x • 104
10−2 = 104−(−2) = 104+2 = 106 → base is 10; used in Scientific Notation
• 10−4 10−3 = 10
−4−(−3) = 10−4+3 = 10−1 = 1
10 → used in Scientific Notation
x−n = 1 xn, x
0 = 1
Example 15. Used to convert a negative exponent into a positive one. • 2−3 = 1
23
• 10−4 = 1 104
(x · y)n= xn· yn
Example 16. The power of a product is the product of the powers. • (2 · 5)3 = 23· 53
• (5x)3 = 53· x3
x y n = x n yn
Example 17. The power of a quotient is the quotient of the powers. • 2 5 2 = 22 52 • x 5 3 = x533 6. Properties of Radicals √ a ·√a = (√a)2 = a, a > 0
Example 18. Taking the square of a square root leaves us with the number inside the radical. We undo the square root by taking the square.
• √5 ·√5 = 5 • (√x)2 = x, x > 0
√
a2 = a, a > 0
Example 19. We can undo the square root by taking the square inside the radical. • √52 = 5 • √x2 = x, x > 0 √ a · b =√a · √ b, a > 0, b > 0
Example 20. The square root of the product is the product of the square roots. • √25 · 16 =√25 ·√16 = 5 · 4 = 20
• px2· y =√x2 ·√y = x√y
r a b = √ a √ b, a > 0, b > 0
Example 21. The square root of the quotient is the quotient of the square roots. • q52 36 = √ 52 √ 36 = 5 6 • qx2 64 = √ x2 √ 64 = x 8 √ a2n = a2n2 = an or √ aeven = aeven2
Example 22. If we have an even exponent inside the radical, 2n above, we can undo the radical by taking a half of the even exponent, 2n2 = n.
• √514= 5142 = 57
• p25x16y10=√25 · x162 · y102 = 5x8y5
• px8y17 = x82 ·py16y = x4· y162 √y = x4y8√y
7. The Slope and Equation of a Line
The slope of a line is a number determined by the coordinates of any two points on the line. If P (x1, y1) and Q(x2, y2) are any two points on the line,
given by their coordinates, then the slope is the number m = y2− y1
x2− x1
= ∆y ∆x
Note that ∆y, the difference in the y-coordinates, is on the top, while ∆x, the difference in the x-coordinates, is on the bottom.
It is a common mistake to use ∆x∆y instead of ∆y∆x to compute the slope.
Notice that we chose above to start with the second point Q and that is why both differences begin with the coordinates of the second point (y2− y1) and (x2 − x1).
One can start instead with the first point P , in which case both differences should start with the coordinates of the first point, namely (y1− y2) and (x1 − x2), which
m = y1− y2 x1− x2 = −(y2− y1) −(x2− x1) = y2− y1 x2− x1
Example 23. Find the slope of the line passing through the points P (−1, −2) and Q(−4, −5). If we choose to start with the first point P , then both differences should start with the coordinates of the first point, with ∆y difference on the top
Slope = m = −2 − (−5) −1 − (−4) = −2 + 5 −1 + 4 = 3 3 = 1
If we choose to start with the second point Q, then both differences should start with the coordinates of the second point, with ∆y difference on the top
Slope = m = −5 − (−2) −4 − (−1) = −5 + 2 −4 + 1 = −3 −3 = 1
Example 24. The following results are useful to remember: • Any horizontal line has slope zero.
• For vertical lines the slope is not defined.
• If a line has a positive slope, the line rises from left to right. • If a line has a negative slope, the line falls down from left to right.
The equation of a line passing through two given points P (x1, y1) and
Q(x2, y2) is given in terms of the slope m and the coordinates of one of the points:
y = y1+ m(x − x1)
Note that y and x are variables, representing the coordinates of an arbitrary point on the line. Here, we chose to use the coordinates of the first point P , namely the given numbers x1 and y1 but one can also use the coordinates of the second point Q
and still get the same equation for the line
Example 25. Take the points above P (−1, −2) and Q(−4, −5), for which we com-puted the slope m = 1. The equation of the line passing through these two points, using the coordinates of the first point, is then
y = −2 + 1(x − (−1)) = −2 + x + 1 = x − 1
We should get the same equation if we use the coordinates of the second point instead y = −5 + 1(x − (−4)) = −5 + x + 4 = x − 1
Example 26. Let the equation of a line be given by 3x + 2y = 5
How can we find the slope of the line from this equation? We need to solve this equation for y in terms of x:
2y = 5 − 3x → subtract from both sides 3x y = 5 − 3x
2 → then divide both sides by 2 y = 5
2 − 3
2x → this is what we need to find the slope Once we have the equation in the form (for some numbers m and b):
y = mx + b → slope = m and y-intercept = b
the slope is simply the number m (including the sign) in front of the variable x, while the number b is the y-intercept. In our case, The slope is the signed number in front of the variable x, namely
slope = −3
2 and the y-intercept = 5 2
A common mistake is to include the variable x in the answer for the slope. Remember that the slope is a number. Another common mistake is to forget the sign and write
3
2 for the slope, instead of − 3
2. Remember that the y-intercept is the y-coordinate of
the point of intersection of the line and the y-axis, when x = 0.
