This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
What don't you like about that problem? The 13 in the denominator, right? Try multiplying both sides by 13 (you'll have to distribute on the left side). Let me know what happens...
For #30 & #32, you want to CLEAR THE DENOMINATORS. The best way to do that is to multiply by the LCD (Least Common Denominator). For #30, the LCD is simply 13... for #32 the LCD is 10.
Learn to ask the question "What don't we like about this problem?" Well, the MOST offensive thing we see right away is that a variable term exists on both sides... WE NO LIKE!
So, we want to GET RID OF one or the other of those terms... what is our "GET RID OF" tool (operation)?... SUBTRACTION, of course!
S0:
4p + 2 = 3p - 7
Let's subtract 3p from both sides... what property are we using? very simply the SUBTRACTION PROP OF =, right?
WHY AM I DOING ALL THE ANSWERING?? (CAPS ARE FOR SHOUTING!!) I am very generous with extra credit for lucid (look it up) blog participation. Gosh, gee willakers... and I just HAPPEN to know a WHOLE BUNCH OF PEOPLE that could use some extra credit, don't I?
Ah-hah!!... That's because you really DON'T have -3 on the left side, you have 5 times -3... you needed to distribute first. We'll take a look at that one tomorrow in class!
How would I do b/13-3b/13=8/13?
ReplyDeleteHelp me! For Homework 2-3?
What don't you like about that problem? The 13 in the denominator, right? Try multiplying both sides by 13 (you'll have to distribute on the left side). Let me know what happens...
ReplyDeletei am really struggling on siplifying. i got all of those qs wrrong on the test. How do you do it?
ReplyDeletei dont get # 26 i got the answer but when i check it its wrongg
ReplyDeletehmwk 2-3 q's # 30 and 32 have completly stumped me! i tried wht you said to the other person but that didnt help me.
ReplyDeleteDear Un-Simple Simon - I need a more specific question... can you give me a specific problem?
ReplyDeleteFor #26, a good first step is to distribute:
ReplyDelete15 = -2(2t - 1)
15 = -2(2t) - (-2)(1)
15 = -4t - (-2)
15 = -4t + 2 ...(now subtract 2 frm bs)
13 = -4t ......(now div bs by -4)
t = -(13/4)
Do you see your error... lmk
For #30 & #32, you want to CLEAR THE DENOMINATORS. The best way to do that is to multiply by the LCD (Least Common Denominator). For #30, the LCD is simply 13... for #32 the LCD is 10.
ReplyDelete#30
b/13 - 3b/13 = 8/13
13( b/13 - 3b/13 ) = (8/13)(13)
13b/13 - 39b/13 = (8*13)/13
b - 3b = 8
-2b = 8
b = -4
I need some of the folks that can answer questions to PARTICIPATE IN THE BLOG AND ANSWER QUESTIONS!!! We are a team!!!
ReplyDeleteI have no idea how to do number 28 !
ReplyDeleteConfuzzled on number 11..
ReplyDeletewhat about my simplifying problem??
ReplyDeleteDear Simplifier - you never gave me a problem to work on... how about a Friday "crunch-lunch" tomorrow? Will I see you there? Bring a friend!
ReplyDeleteRiddle: What's better that 80 minutes of algebra?
Answer: 120 minutes!
here's a problem similar to #28:
ReplyDelete3x + 2(2x-3) = 64 ... distribute
3x + 4x - 6 = 64 ... combine like terms
7x - 6 = 64 ... add prop of =
7x = 70 ... div prop of =
x = 10 ... voila SOLUTION!
don't forget to check!
For #11...
ReplyDeleteLearn to ask the question "What don't we like about this problem?" Well, the MOST offensive thing we see right away is that a variable term exists on both sides... WE NO LIKE!
So, we want to GET RID OF one or the other of those terms... what is our "GET RID OF" tool (operation)?... SUBTRACTION, of course!
S0:
4p + 2 = 3p - 7
Let's subtract 3p from both sides... what property are we using? very simply the SUBTRACTION PROP OF =, right?
So now we have:
4p - 3p + 2 = 3p - 3p - 7 ... simplify to...
p + 2 = -7
I'm hoping that you can take it from here.
WHY AM I DOING ALL THE ANSWERING?? (CAPS ARE FOR SHOUTING!!) I am very generous with extra credit for lucid (look it up) blog participation. Gosh, gee willakers... and I just HAPPEN to know a WHOLE BUNCH OF PEOPLE that could use some extra credit, don't I?
ReplyDeleteI'm having problems with #22
ReplyDelete5(2x-3)=15
+3 +3
5(2x)= 18
---- --
2 2
5/2.x=9
--- ---
5/2 5/2
x =18/5
But I know the answer has to be 3. HELP
you said something in class about retakes?
ReplyDeleteAh-hah!!... That's because you really DON'T have -3 on the left side, you have 5 times -3... you needed to distribute first. We'll take a look at that one tomorrow in class!
ReplyDelete# 26. I keep getting crazy numbers like 44/-8 and I don't know what to do. Help!
ReplyDeleteDear Crazy,
ReplyDeleteWhat was your first step for #26? btw, it is an ugly answer -13/4 or -3.25, but I kinda think there's some beauty in it!
On witch page ... if it is pg 98 or pg 106 i can help....
ReplyDeleteTELL ME WOTCH ONE IT IS
Regarding number 26
ReplyDeleteYou are completely right. It like beauty and the beast!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ReplyDeleteMr Chamberlain with you make a topic for Quiz Questions?
ReplyDelete-Lotta