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Why is Algebra so hard for me
[#permalink]
29 Jan 2011, 12:28
29 Jan 2011, 12:28
I have just started studying for my GMAT in the last couple days and Algebra makes my eyes glaze over. I know I need to master the basic principles of Algebra but I am having such a hard time.
I keep looking over and over at the basic principles that were compiled on the site and I think my brain only can see the connection after seeing several examples of a principle.
_____________________________
The sum of N consecutive integers:
S=(n+n+N-1)*N/2=n*N+N*(N-1)/2
S/N=n+(N-1)/2
If N is odd, then S would be a multiple of N. If N is even, then S would not be a multiple of N.
Can anyone give an example of this?
____________________________
To find the GCF/LCM, you will need to do prime-factorization. This means reducing a number to its prime-factor form.
E.g. 1
GCF/LCM of 4,18
4 = 2*2
18= 2*3*3
To find the GCF, take the multiplication of the common factors. In this case, GCF = 2.
To find the LCM, take the multiplication of all the factors. In this case, LCM=2*2*3*3=36
E.g. 2
GCF/LCM of 4,24
4 = 2*2
24 = 2*2*2*3
GCF = 2*2 = 4 (Why is this GCF 4 when in EG 1 the GCF is 2?)
LCM = 2*2*2*3 = 24
____________________________
If anyone has examples on the following principles that would help me out alot!
1)(n!)^2 > n^n
Inequalities
2)You need to flip signs when 1 is divided by both side:
1/a<1/b, 1/c<1/d
3)Deal with negative numbers:
-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0 How does this Convert into this? -a-(-d)<-b-(-c)
Cancelling Common Terms
Example:
x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
How does -x all of a sudden turn into -1? How would this example fit into a GMAT question?
I have more questions but dont have enough time to post them. I feel that there are so many assumptions that I am suppose to know when math is suppose to be pretty straight forward!!
Thanks for the help
I keep looking over and over at the basic principles that were compiled on the site and I think my brain only can see the connection after seeing several examples of a principle.
_____________________________
The sum of N consecutive integers:
S=(n+n+N-1)*N/2=n*N+N*(N-1)/2
S/N=n+(N-1)/2
If N is odd, then S would be a multiple of N. If N is even, then S would not be a multiple of N.
Can anyone give an example of this?
____________________________
To find the GCF/LCM, you will need to do prime-factorization. This means reducing a number to its prime-factor form.
E.g. 1
GCF/LCM of 4,18
4 = 2*2
18= 2*3*3
To find the GCF, take the multiplication of the common factors. In this case, GCF = 2.
To find the LCM, take the multiplication of all the factors. In this case, LCM=2*2*3*3=36
E.g. 2
GCF/LCM of 4,24
4 = 2*2
24 = 2*2*2*3
GCF = 2*2 = 4 (Why is this GCF 4 when in EG 1 the GCF is 2?)
LCM = 2*2*2*3 = 24
____________________________
If anyone has examples on the following principles that would help me out alot!
1)(n!)^2 > n^n
Inequalities
2)You need to flip signs when 1 is divided by both side:
1/a<1/b, 1/c<1/d
3)Deal with negative numbers:
-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0 How does this Convert into this? -a-(-d)<-b-(-c)
Cancelling Common Terms
Example:
x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
How does -x all of a sudden turn into -1? How would this example fit into a GMAT question?
I have more questions but dont have enough time to post them. I feel that there are so many assumptions that I am suppose to know when math is suppose to be pretty straight forward!!
Thanks for the help
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Re: Why is Algebra so hard for me
[#permalink]
29 Jan 2011, 14:27
29 Jan 2011, 14:27
1. Consecutive integers
For instance, you have a set of numbers: 4,5,6. The sum is 15, it is divisible by 3 (we have 3 numbers). Or consider this set: 5,6,7,8,9. The sum is 35, it's divisible by 5.
Now consider these sets: 4,5,6,7. The sum is 22, it's not divisible by 4. Or 5,6,7,8. S=26, again, not divisible by 4.
2. Prime-factorization.
In the first example the only common factor of 4 and 18 is 2^1. ( it's a factor of BOTH 4 and 18)
In the second example common factor of 4 and 24 is 2^2 (BOTH 4 and 24 are divisible by 4, we need to find maximum factor which divides evenly into both numbers)
Venn diagrams can be helpful there. For instance, we have 20 and 35. Draw 2 cirles, in one we have 2,2,5 (these are factors of 20), in other we have 5,7 (factors of 35). The only common factor is 5 - that's where our circles overlap. To find LCF we simply get ALL factors of BOTH numbers - 2,2,5,7. That's 140.
3. n! means factorial - the product of all numbers lower than n and n. For example 4!=1*2*3*4=24. 24^2 =576. Now, 4^4 (n=4 in our case) equals 256, that is less than 576. The bigger n we get, the bigger gap we have.
4. For example, we have 2 and 3. 2 is clearly less than 3, however, when we divide by 1 we change the sign: 1/2>1/3 (since we divide by LESSer amount). Always look at the signs though (whether numbers are positive or negative)
5. "-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0" Hm, didn't get this part but since you are asking another issue, consider this example:
-2-5<-1-3, now, we are carrying over -5 and -3: -2-(-3)<-1-(-5) => 1<4. When we are doing this we must change the sign as we did in this example. Also notice that subtracting negative number equals to adding positive number.
6. Cancelling common terms. Actually you can meet similar equations in the real exam, that's a 500-600 level question I suppose. I've seen similar problems in OG. Firstly, we cannot divide by x because by doing that we can incorrectly cross out one of the solutions (0 in our example). After moving x to the left side we can notice that x is a common factor of x(x-2) and x, i.e. both of these are divisible by x. So we can factor x out: x(x-2-1). When we divide x(x-2) by x we get (x-2) and when we divide x by x we get 1, minus remains. That's it I believe.
If I was wrong somewhere, correct me.
For instance, you have a set of numbers: 4,5,6. The sum is 15, it is divisible by 3 (we have 3 numbers). Or consider this set: 5,6,7,8,9. The sum is 35, it's divisible by 5.
Now consider these sets: 4,5,6,7. The sum is 22, it's not divisible by 4. Or 5,6,7,8. S=26, again, not divisible by 4.
2. Prime-factorization.
In the first example the only common factor of 4 and 18 is 2^1. ( it's a factor of BOTH 4 and 18)
In the second example common factor of 4 and 24 is 2^2 (BOTH 4 and 24 are divisible by 4, we need to find maximum factor which divides evenly into both numbers)
Venn diagrams can be helpful there. For instance, we have 20 and 35. Draw 2 cirles, in one we have 2,2,5 (these are factors of 20), in other we have 5,7 (factors of 35). The only common factor is 5 - that's where our circles overlap. To find LCF we simply get ALL factors of BOTH numbers - 2,2,5,7. That's 140.
3. n! means factorial - the product of all numbers lower than n and n. For example 4!=1*2*3*4=24. 24^2 =576. Now, 4^4 (n=4 in our case) equals 256, that is less than 576. The bigger n we get, the bigger gap we have.
4. For example, we have 2 and 3. 2 is clearly less than 3, however, when we divide by 1 we change the sign: 1/2>1/3 (since we divide by LESSer amount). Always look at the signs though (whether numbers are positive or negative)
5. "-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0" Hm, didn't get this part but since you are asking another issue, consider this example:
-2-5<-1-3, now, we are carrying over -5 and -3: -2-(-3)<-1-(-5) => 1<4. When we are doing this we must change the sign as we did in this example. Also notice that subtracting negative number equals to adding positive number.
6. Cancelling common terms. Actually you can meet similar equations in the real exam, that's a 500-600 level question I suppose. I've seen similar problems in OG. Firstly, we cannot divide by x because by doing that we can incorrectly cross out one of the solutions (0 in our example). After moving x to the left side we can notice that x is a common factor of x(x-2) and x, i.e. both of these are divisible by x. So we can factor x out: x(x-2-1). When we divide x(x-2) by x we get (x-2) and when we divide x by x we get 1, minus remains. That's it I believe.
If I was wrong somewhere, correct me.
Re: Why is Algebra so hard for me
[#permalink]
29 Jan 2011, 17:25
29 Jan 2011, 17:25
DaoEmo wrote:
1. Consecutive integers
For instance, you have a set of numbers: 4,5,6. The sum is 15, it is divisible by 3 (we have 3 numbers). Or consider this set: 5,6,7,8,9. The sum is 35, it's divisible by 5.
Now consider these sets: 4,5,6,7. The sum is 22, it's not divisible by 4. Or 5,6,7,8. S=26, again, not divisible by 4.
2. Prime-factorization.
In the first example the only common factor of 4 and 18 is 2^1. ( it's a factor of BOTH 4 and 18)
In the second example common factor of 4 and 24 is 2^2 (BOTH 4 and 24 are divisible by 4, we need to find maximum factor which divides evenly into both numbers)
Venn diagrams can be helpful there. For instance, we have 20 and 35. Draw 2 cirles, in one we have 2,2,5 (these are factors of 20), in other we have 5,7 (factors of 35). The only common factor is 5 - that's where our circles overlap. To find LCF we simply get ALL factors of BOTH numbers - 2,2,5,7. That's 140.
3. n! means factorial - the product of all numbers lower than n and n. For example 4!=1*2*3*4=24. 24^2 =576. Now, 4^4 (n=4 in our case) equals 256, that is less than 576. The bigger n we get, the bigger gap we have.
4. For example, we have 2 and 3. 2 is clearly less than 3, however, when we divide by 1 we change the sign: 1/2>1/3 (since we divide by LESSer amount). Always look at the signs though (whether numbers are positive or negative)
5. "-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0" Hm, didn't get this part but since you are asking another issue, consider this example:
-2-5<-1-3, now, we are carrying over -5 and -3: -2-(-3)<-1-(-5) => 1<4. When we are doing this we must change the sign as we did in this example. Also notice that subtracting negative number equals to adding positive number.
6. Cancelling common terms. Actually you can meet similar equations in the real exam, that's a 500-600 level question I suppose. I've seen similar problems in OG. Firstly, we cannot divide by x because by doing that we can incorrectly cross out one of the solutions (0 in our example). After moving x to the left side we can notice that x is a common factor of x(x-2) and x, i.e. both of these are divisible by x. So we can factor x out: x(x-2-1). When we divide x(x-2) by x we get (x-2) and when we divide x by x we get 1, minus remains. That's it I believe.
If I was wrong somewhere, correct me.
For instance, you have a set of numbers: 4,5,6. The sum is 15, it is divisible by 3 (we have 3 numbers). Or consider this set: 5,6,7,8,9. The sum is 35, it's divisible by 5.
Now consider these sets: 4,5,6,7. The sum is 22, it's not divisible by 4. Or 5,6,7,8. S=26, again, not divisible by 4.
2. Prime-factorization.
In the first example the only common factor of 4 and 18 is 2^1. ( it's a factor of BOTH 4 and 18)
In the second example common factor of 4 and 24 is 2^2 (BOTH 4 and 24 are divisible by 4, we need to find maximum factor which divides evenly into both numbers)
Venn diagrams can be helpful there. For instance, we have 20 and 35. Draw 2 cirles, in one we have 2,2,5 (these are factors of 20), in other we have 5,7 (factors of 35). The only common factor is 5 - that's where our circles overlap. To find LCF we simply get ALL factors of BOTH numbers - 2,2,5,7. That's 140.
3. n! means factorial - the product of all numbers lower than n and n. For example 4!=1*2*3*4=24. 24^2 =576. Now, 4^4 (n=4 in our case) equals 256, that is less than 576. The bigger n we get, the bigger gap we have.
4. For example, we have 2 and 3. 2 is clearly less than 3, however, when we divide by 1 we change the sign: 1/2>1/3 (since we divide by LESSer amount). Always look at the signs though (whether numbers are positive or negative)
5. "-a<-b<0, -c<-d<0
Then
-a-c<-b-d<0" Hm, didn't get this part but since you are asking another issue, consider this example:
-2-5<-1-3, now, we are carrying over -5 and -3: -2-(-3)<-1-(-5) => 1<4. When we are doing this we must change the sign as we did in this example. Also notice that subtracting negative number equals to adding positive number.
6. Cancelling common terms. Actually you can meet similar equations in the real exam, that's a 500-600 level question I suppose. I've seen similar problems in OG. Firstly, we cannot divide by x because by doing that we can incorrectly cross out one of the solutions (0 in our example). After moving x to the left side we can notice that x is a common factor of x(x-2) and x, i.e. both of these are divisible by x. So we can factor x out: x(x-2-1). When we divide x(x-2) by x we get (x-2) and when we divide x by x we get 1, minus remains. That's it I believe.
If I was wrong somewhere, correct me.
I really appreciate you taking the time to explain these. All of these were from the Basic Math Principles on the forum. I understand all the concepts except for #6
