# I suck at math...again

Registered User, __BANNED USERS regular
edited April 2009
I do not understand why this is not working. It is another one of those fucking sally has 30,000 invested and received 2,300 interest. She invested at 10 percent and 5 percent how much did she invest at each rate.

x+y= 30,000
y=30,000-x
(.1x+.05y) 30000=2,300

(.1x + .05[30000-x]) 30000
3000x+45000000-1500x
1500x=-44997700
x=-29998.5

That is obviously not right, what the hell did I do wrong? This is so frustrating because everyone else in the class is saying how bad they are at math and how tough this is yet they fly by all this shit and I am stuck on the easy stuff like this. I did it exactly how the book says to and I still can not get it right after an hour of work. Fuck sally.

Fizban140 on

## Posts

• Registered User regular
edited March 2009
0.1x + .05y = 2300
x+y = 30,000

If you want the answer (I think):

Simultaneous equations.
Re arrange to

x = 30,000 - y
Substitute into the other equation.

Get

0.05y + 3000 -0.1y = 2300

y = 14000
therefore x = 16000

Teslan26 on
• Registered User, __BANNED USERS regular
edited March 2009
Fuck...god damnit math is fucking frustrating. Wasting all my free time because I got stuck on the simplest step.

Fizban140 on
• Registered User regular
edited March 2009
Teslan26 wrote: »
0.1x + .05y = 2300
x+y = 30,000

This is what I was going to write.

Do you understand why what you have is wrong and why the above is correct though? We can help you reason through it if you need help.

Al_wat on
• Registered User regular
edited March 2009
Put full answer in spoiler above

Teslan26 on
• Registered User, __BANNED USERS regular
edited March 2009
I messed up because .1 and .5 times x and y is the interest not the principle. Right?

Fizban140 on
• Registered User regular
edited March 2009
I don't understand what you just said. Sorry.

x the number of shares bought at rate 10% interest
y the number of shares bought at rate 5% interest

I know the total number of shares: x+y = 30 000
I know the total interest, since this is only for 1 year (I assumed) I simply multiplied the number of dollars by the relevant interest rate. Then added them to get 2300.

Teslan26 on
• Registered User, __BANNED USERS regular
edited March 2009
This is suprising, I got stuck on the very next question for probably the same reason.

Tickets for a concert were sold to adults for $3 and students for$2 if the total receipts were $824 and twice as many adult tickets were sold as student then how many of each were sold? Fizban140 on • Registered User regular edited March 2009 x = 2y x = adult tickets y = child tickets Yes? See if you can see the second equation 3x + 2y = 824 Teslan26 on • Registered User regular edited March 2009 So the solution is: 3*(2y) + 2y = 824 8y = 824 y =103 x will therefore = 206 Agreed? What is sick, is that I am actually enjoying doing this..... Teslan26 on • Registered User, __BANNED USERS regular edited March 2009 All I was able to get for the past 10 minutes was 3x+2y=824 how does x = 2y help anything? I do not understand this, I probably should not be taking this math class. Fizban140 on • Registered User regular edited March 2009 There are two options here, I will outline both. Knowing x = 2y, you can simply replace all instances of 'x' with '2y' as above 3x+2y=824 becomes 3*(2y) + 2y = 824 Then as above 8y = 824 y =103 x will therefore = 206 OR x = 2y ---> x - 2y = 0 x - 2y = 0 3x+2y=824 Add the two equations, the 'y' cancels, and we have 4x = 824 Which method have you been shown in class? Teslan26 on • Registered User, __BANNED USERS regular edited March 2009 I just got it, I am not sure which method we were shown in class. The teacher is about 70 years old and I can not pay attention to him at all, I know it sounds horrible but the class is about 85 to 90 degrees and he speaks incredibly slowly. I try but he puts me to sleep so I try and work on problems while I listen to him. Fizban140 on • Registered User regular edited March 2009 Fizban140 wrote: » All I was able to get for the past 10 minutes was 3x+2y=824 how does x = 2y help anything? I do not understand this, I probably should not be taking this math class. you input x = 2y into your first equation. The way we would do it in my uni math classes (which I have very nearly completely forgotten) is like this: Equation (1): 3x + 2y = 824 Equation (2): x = 2y Plug (2) into (1) 3 (2y) + 2y = 824 (from here reduce, I'm not going to run through it since Teslan already did). I swear I keep trying to write explanations of my reasoning but I feel they are too complicated (I'm going off pure intuition here - like I said I barely remember my math courses). Al_wat on • Smells great! Houston, TXRegistered User regular edited March 2009 Fizban140 wrote: » All I was able to get for the past 10 minutes was 3x+2y=824 how does x = 2y help anything? I do not understand this, I probably should not be taking this math class. Keep in mind that, when you have to solve an equation with multiple variables, the only way to solve them is to have the same number of unique equations as you have variables. If you have one variable, you only need one equation to solve it. If you have two variables, such as in your example above (number of adult tickets and number of child tickets), then you need two equations to solve for them. You understand the first equation, which is 3x + 2y = 824. Three bucks times the number of adult tickets plus two bucks times the number of student tickets gives the total, which is$824. So you need a second equation now to solve for both variables. So think, where can you get another unique equation from the problem?

The answer is in the second part of the problem, where it says that twice as many adult tickets as student tickets. That information alone is enough to give you a whole new equation, which is x = 2y... in other words, the number of adult tickets equals two times the number of child tickets. Now you have two variables and two equations, and can solve the system.

Big Dookie on
Steam | Twitch
Oculus: TheBigDookie | XBL: Dook | NNID: BigDookie
• Registered User, __BANNED USERS regular
edited March 2009
Big Dookie wrote: »
Fizban140 wrote: »
All I was able to get for the past 10 minutes was 3x+2y=824

how does x = 2y help anything? I do not understand this, I probably should not be taking this math class.
Keep in mind that, when you have to solve an equation with multiple variables, the only way to solve them is to have the same number of unique equations as you have variables. If you have one variable, you only need one equation to solve it. If you have two variables, such as in your example above (number of adult tickets and number of child tickets), then you need two equations to solve for them.

You understand the first equation, which is 3x + 2y = 824. Three bucks times the number of adult tickets plus two bucks times the number of student tickets gives the total, which is $824. So you need a second equation now to solve for both variables. So think, where can you get another unique equation from the problem? The answer is in the second part of the problem, where it says that twice as many adult tickets as student tickets. That information alone is enough to give you a whole new equation, which is x = 2y... in other words, the number of adult tickets equals two times the number of child tickets. Now you have two variables and two equations, and can solve the system. I wish things were explained to us like this in either the book or by the teacher. Instead he just reads examples out of the book and works it out on the board for two and a half hours. Fizban140 on • Registered User regular edited March 2009 It is nearly 2am - so I am gonna let these fine gentlemen handle any further issues. Simultaneous equations are all pretty similar though, once the trick works in your head - it is easy for life. Challenge, for when you feel confident with it: If you can do it with 2, can you do it with 3? x +y + z = 33 2x - y - 2z = -7 4x + y + 4z = 27 Hint: Find one variable/letter in terms of everything else. Sub that into the two unused equations, and you have a simple 2 equation problem I fully accept you might not bother to do this. It is a good aid to understanding the idea behind it all though. Teslan26 on • Registered User regular edited March 2009 Fizban140 wrote: » All I was able to get for the past 10 minutes was 3x+2y=824 how does x = 2y help anything? x is the number of adult tickets sold and y is the number of student tickets sold. Since the problem says that twice the number of adult tickets were sold as student tickets, then we know that x is the same number as 2y. Writing x as 2y helps because it lets us rewrite the equation in terms of one variable, making it easier to solve. 3x + 2y = 824 is the same as 3(2y) + 2y = 824, which is the same as 6y + 2y = 824 which is just 8y = 824. That last equation should be easy to solve. Fizban140 wrote: » I do not understand this, I probably should not be taking this math class. I always hated math and told myself that I wasn't good at it. This past year I've taken Introduction to Calculus at university. I'm averaging a 50 in it, but if I've learned one thing, it's that learning to do math is like learning to play an instrument/juggle/draw/whatever you've been taught in basic training. It's not about having some mystical talent, it's about the hours you put in to it. Telling yourself that you don't understand what you're doing and shouldn't even be trying isn't going to help you do this. A defeatist attitude helps no one. You're in the army, right? Since when do they teach you to give up when faced with a challenge? You can do this stuff. Klorgnum on • Registered User, __BANNED USERS regular edited March 2009 I might be here all night, moved on to the very next question after that and I have been stuck since. -3x + 5y = 1 9x - 3y = 5 I have about 2 pages of work and I keep getting y-.25 and x= .08 but that does not work. I have worked it out like the book says but it will not work. Fizban140 on • Registered User regular edited March 2009 Fizban140 wrote: » I might be here all night, moved on to the very next question after that and I have been stuck since. -3x + 5y = 1 9x - 3y = 5 I have about 2 pages of work and I keep getting y-.25 and x= .08 but that does not work. I have worked it out like the book says but it will not work. The first thing I notice here is that 3 is a factor of 9. So if we multiply the top equation by 3... 3(-3x + 5y) = 3(1) we now have -9x + 15y = 3 -9x is just the opposite of 9x, so we can add the two equations together and get rid of one of the variables. We just add the left hand sides and the right hand sides together. -9x + 15y = 3 9x - 3y = 5 -9x + 15y + 9x - 3 y = 3 + 5 which is the same as 12y = 8 (because adding -9x to 9x gives 0, so we have no more x's in this equation and can solve for y) y = 8/12 y = 2/3 = 0.66666666... Now that we have y = 2/3 we can just plug that into the first equation and solve for x. Klorgnum on • Registered User regular edited March 2009 First step is to just solve one of the equations for either x or y, I choose y. Next step is to plug that into the second equation. That will get you one of the variables. Then you can plug that into what you found to be the relation between the variables. -3x + 5y = 1 5y = 3x + 1 y = 3/5 x + 1/5 9x - 3y = 5 9x - 3(3/5 x + 1/5) = 5 9x - 9/5 x - 3/5 = 5 36/5 x = 28/5 x = 28/36 = 79 y = 3/5 x + 1/5 y = 3/5 (7/9) + 1/5 y = 7/15 + 1/5 y = 2/3 EDIT: Or you can do what Klorgnum did above as he did some fancy multiplication to deal with a lot less fractions, and is probably ultimately the easier way. Although this way can be done every time regardless of whether you can cancel out equations. Institutional Pederasty on • Registered User regular edited March 2009 First step is to just solve one of the equations for either x or y, I choose y. Next step is to plug that into the second equation. That will get you one of the variables. Then you can plug that into what you found to be the relation between the variables. -3x + 5y = 1 5y = 3x + 1 y = 3/5 x + 1/5 9x - 3y = 5 9x - 3(3/5 x + 1/5) = 5 9x - 9/5 x - 3/5 = 5 36/5 x = 28/5 x = 28/36 = 79 y = 3/5 x + 1/5 y = 3/5 (7/9) + 1/5 y = 7/15 + 1/5 y = 2/3 EDIT: Or you can do what Klorgnum did above as he did some fancy multiplication to deal with a lot less fractions, and is probably ultimately the easier way. Although this way can be done every time regardless of whether you can cancel out equations. What it really comes down to is knowing which method to use when. When you see the question, ask yourself if there's anything you can easily do to get rid of one of the variables. If there is, do it my way. If there isn't, then go for Institutionalised Pederasty's method. It seems likely that the problem would have been written with one specific method in mind, so if one way is giving you a messy looking equation, try another. I'd be willing to bet that the correct answer is going to look neat and tidy 75% of the time. Also, once you know how to get the answer with one method, try solving it with another for the same results. If you can do it correctly more than one way, you'll have an easier time solving similar equations. My method isn't going to be the easiest one all the time, and neither is Institutionalised Pederasty's, but if one of them is hard to use, there's a good possibility that the other might be fairly easy. Klorgnum on • Registered User, __BANNED USERS regular edited March 2009 3/2x - 2y/3 = 10 1/2 x - 1/2 y = -1 The obvious method would be to multiply by the LCD but I am getting some ridiculous numbers that I have no clue on how to work with. I got x=9/13 which can not be right and I have no idea how that is suppose to be thrown into another equation and still make sense. Fizban140 on • Registered User regular edited March 2009 I'm really liking the look of the second equation so I'll start there. Now besides multiplying the bottom equation by 2, I used the same method as before to solve this system. 2 (1/2 x - 1/2 y = -1) x - y = -2 x = y - 2 3/2 (y-2) - 2y/3 = 10 3/2 y - 3 - 2/3 y = 10 3/2 y - 2/3 y = 13 5/6 y = 13 y = 78/5 x = y - 2 x = 78/5 -2 x = 68/5 Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 I am way off I just got x-4 and y=-6, guess I have to start all over then. I have been working on this shit for 4 hours and I am not even 1/3rd the way done with my homework. Is this wrong? 9x - 4 y = 60 (I multiplied by 6) x+y = -2 (then multiply by 4) 4x+4y=-8 13x=52 x=52/13 Fizban140 on • Registered User regular edited March 2009 Should be x - y = -2 and therefore 4x - 4y = -8 Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 I gave up on that one and ended with x= 52/13 and y=0 probably not right but I am going to be up all night at this rate and these are the easy problems. Fizban140 on • Registered User regular edited March 2009 The thing is you were so very close on that one. You had the idea correct. You just weren't careful with your signs and switched a negative for a positive. Mainly, what I'm trying to say is, Don't get so down on yourself, you had the idea on that one and show that you know what you are doing. You just need to take it easy and be careful when dealing with signs, something that plagues me even today. Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 2y-x=3 x=3y-5 I keep getting y=2 and x=-11 which is not right. I am doing 2y-(3y-5)=3 2y-3y+5=3 y=-2 x=-6-5 x=11 Is does not work when I plug it in though, why not? Fizban140 on • Registered User regular edited March 2009 Fizban140 wrote: » I am doing 2y-(3y-5)=3 2y-3y+5=3 y=-2 Again the signs are getting you. You should get -y = -2 y = 2 x = 3y -5 x = 6 - 5 x = 1 Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 What am I doing wrong here? Besides having to ask for help on every question of course... 1/2 the boys and 1/3 the girls attended the game whereas 1/3 the boys and 1/2 the girls attended the dance. if there were 570 students at the game and 580 at the dance then how many are there at freemont high? 1/2x + 1/3y = 570 1/3x + 1/2y = 580 3x + 2y = 3420 2x + 3y = 3480 5x + 5y = 6,700 x + y = 268 Obviously that is already wrong, what did I do wrong? Fizban140 on • Registered User regular edited March 2009 Fizban140 wrote: » 3x + 2y = 3420 2x + 3y = 3480 5x + 5y = 6,700 x + y = 268 Obviously that is already wrong, what did I do wrong? Well firstly, 3420 + 3480 = 6900, not 6700. Secondly, for the next step what did you do? You should divide through by 5, which would yield: x + y = 1380. Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 I somehow got that one, there is probably some rule I was never taught in class that I have to follow and eventually by doing it twenty times I got it but now I am stuck on the next one. I can make x=1 z=2 work for 2 of the equations and then y=-2 works for the first equation but not the other two...what the fuck is wrong? Fizban140 on • Registered User regular edited March 2009 Fizban140 wrote: » I somehow got that one, there is probably some rule I was never taught in class that I have to follow and eventually by doing it twenty times I got it but now I am stuck on the next one. I can make x=1 z=2 work for 2 of the equations and then y=-2 works for the first equation but not the other two...what the fuck is wrong? Obviously you did something wrong with your algebra. Try it again. If it doesn't work for all three equations, well you did something wrong. Demerdar on • Registered User, __BANNED USERS regular edited March 2009 I keep getting stuck on the last variable, I know the first two are right but the last one keeps changing depending upon which of the three equations I use it in. Nevermind I am not even close to the answer, the problem is nothing like the example and I get a different variable every single time. Here is a fun one x-y=3 y+z=8 2x+2z=7 As far as I can tell this is an inconsistent equation. Fizban140 on • Registered User regular edited March 2009 x-y=3 x = y + 3 y + z = 8 z = -y + 8 2(y + 3) + 2(-y + 8) = 7 2y + 6 - 2y + 16 = 7 22 = 7 Ok, something weird is happening in this one. Institutional Pederasty on • Registered User, __BANNED USERS regular edited March 2009 If I am struggling this much with college algebra is it even worth going on any further? I have spent about eight hours on homework that shouldn't take slow people more than two. Eventually it is going to become unmanageable and even if I am able to do it if I am so inefficient at something what company would ever want to hire me? I am giving up for the night after skipping three problems I can not solve, they are all like this. Harry has 2.25 in nickels, dimes and querters. If he had twice as many nickels, half as many dimes and the same amount of quarters he would ave 2.50, If he has 27 counds altogether then how many of each does he have? Fizban140 on • Registered User regular edited March 2009 x-y=3 x = y + 3 y + z = 8 z = -y + 8 2(y + 3) + 2(-y + 8) = 7 2y + 6 - 2y + 16 = 7 22 = 7 Ok, something weird is happening in this one. Um.. what? I can't follow your work, like at all. Demerdar on • Registered User regular edited March 2009 Fizban140 wrote: » If I am struggling this much with college algebra is it even worth going on any further? I have spent about eight hours on homework that shouldn't take slow people more than two. Eventually it is going to become unmanageable and even if I am able to do it if I am so inefficient at something what company would ever want to hire me? First off, have some confidence in yourself. You got the idea in time. You are capable. Second, signs are vital. You will spend bloody ages going through your own work checking signs, I make mistakes in that all the time (it is a running joke that my friends at uni caught their sign screw ups from me) and learning to look for your own errors is a good skill. Elementary errors =/= stupidity! Finally, if you cannot pay attention in class, then you will not get this stuff without tutoring. I'll bet you I could have wrapped this topic up for you in about 30 mins of face time. A forum makes it all that much slower and more difficult. Even at my most lazy, I would skip homework but always show up and pay attention in class. That is where the knowledge is at. That is the place you'll learn. Sit at the front. Copy down the examples perfectly. Make sure you understand everything you write down, if you don't then ask and force the explanation. There are lots of bad teachers out there, and mathematics suffers greatly under them, but there are also bad students. Teslan26 on • Registered User regular edited March 2009 Fizban140 wrote: » I keep getting stuck on the last variable, I know the first two are right but the last one keeps changing depending upon which of the three equations I use it in. Nevermind I am not even close to the answer, the problem is nothing like the example and I get a different variable every single time. Here is a fun one x-y=3 y+z=8 2x+2z=7 As far as I can tell this is an inconsistent equation. That seemingly does not work. I have 22=7 as well, by a different method to IP. ^_^ Teslan26 on • Registered User regular edited March 2009 Fizban140 wrote: » If I am struggling this much with college algebra is it even worth going on any further? I have spent about eight hours on homework that shouldn't take slow people more than two. Eventually it is going to become unmanageable and even if I am able to do it if I am so inefficient at something what company would ever want to hire me? I am giving up for the night after skipping three problems I can not solve, they are all like this. Harry has 2.25 in nickels, dimes and querters. If he had twice as many nickels, half as many dimes and the same amount of quarters he would ave 2.50, If he has 27 counds altogether then how many of each does he have? Okay, well here are the two easy equations: n = # of nickles d = # of dimes q = # of quarters n + d + q = 27 (the sum of all the coins = 27) .05n + 0.10d + 0.25q = 2.25 (he has a total$2.25 in his pocket. Nickles are worth 5 cents, or 5% of a dollar. Same with quarters and dimes. The sum of these is equal to the amount of money in your pocket)

Now here is the last one:

0.1n + 0.05d + 0.25q = 2.50

"If he had twice as many nickels, half as many dimes and the same amount of quarters he would ave $2.50" So we can use this n = 2n (TWICE as many nickles as before) d = 0.5d (HALF as many dimes as before) q = q (SAME amount of quarters before) Now all you do is plug your new n in for your old n, d for your old d, and q for your old q (but this q doesn't change because q=q right?) So it follows: 0.05(2n) + 0.10(0.5d) + 0.25q = 2.50 (because now you have$2.50 in your pocket.

0.10n + 0.05d + 0.25q = 2.50

The reason we can make this substitution because we have a constant total number of coins, 27. But we can change how many of each coin we have. In this case we have twice as many nickles, which ups their overall value. Less dimes than before, so they are weighted less out of our $2.50. Understand? If you solve it all out you get 15 nickles (75 cents) 10 dimes ($1.00)
2 quarters (50 cents)

We have 27 coins and \$2.25 in our pocket. Yay!

Demerdar on