Khargosh Posted January 30, 2007 Share Posted January 30, 2007 Hello, I'm a junior in Highschool, currently taking Algebra II. The reason I'm making this topic, is due to the many problem's i often find myself facing, when trying to work out problem's. Previous to this class, I've taken two classes of basic Algebra and one course in basic geometry, and to this day, comprehend very little of it. At this point in time in the class, we are reviewing the algebra 1 material, and i am at a big loss. We had a practice test today, covering the materials so far, and out of 25, i missed at most 12. For the most part, the way to solve the problem's is what trip's me up, and i often spend loads's of time picking number's from my head, and plugging them into the equation. I get the right answer by doing so, but it is very time consuming. I've written a few problem's from my class, that went over my head completley, and will write them down, along with how and what i did, perhaps some of you can provide some much needed and appreciated help. I also am trying, to once and for all tackle this algebra so that i may possibly take chemistry/physics courses, both of which i am told require algebra related math,and difficult ones at that. So another question i have, is if i do take one of said classes, what kind of math can i expect to see? and for someone in my situation would it be wise to take these classes? I'll get to the question's now. SOLVE AND GRAPH 1)3(4d-9)<6(2d-5) for this one i distributed the 3 to both the 4d and -9 and got... 12d-27 and then distributed the 6 to the 2d and -5 and got... 12d-30 so i end up with 12d-27>12d-30 my question is, where to go from there? 2) |4x-3|=-27 I was told that the two lines (absolute value?, mean two equations) so i ended up 4x-3=-27 and 4x-3=27 question being do i use either [and/or] and where to go from there. 3) 3(4x+7)<21 i distribute 3 to 4x and 7 and get 12x+21<21 from here where do i go? 4) This one i'm not sure how to approach, if someone could give me a tip on where to start. 2x+3/5<0.03 these are the one's i had the msot trouble with, and would greatly apprciate the help. -Thanks Brendan Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted January 30, 2007 Share Posted January 30, 2007 2) |4x-3|=-27 I was told that the two lines (absolute value?, mean two equations) so i ended up Hold it. That's not what absolute value means. Absolute value means "this number, and if it's negative, make it positive," to put it simply. That means that [math]|-4| = 4[/math] and [math]|3| = 3[/math] So pause and look at that a moment. If you take the absolute value of an expression, the result is always positive. That means that [math]|4x-3|[/math] cannot be equal to -27, ever. So there's no solution. Link to comment Share on other sites More sharing options...

EvoN1020v Posted January 30, 2007 Share Posted January 30, 2007 Be careful: Absolute value actually means the distance between Object A and Object B regardless of their directions. Link to comment Share on other sites More sharing options...

the tree Posted January 30, 2007 Share Posted January 30, 2007 Question 1 makes no sense, it basically states that -27<-30 which is obviously not true, regardless of values of d. Question 2 is equally nonsensical as Cap'n explained. Question 3 [math]12x+21\le 21[/math] [math]12x\le 0[/math] [math]x\le 0[/math] ^ graph that. Question 4 [math]2x+\frac{3}{5}< 0.03[/math] [math]2x + 0.6 < 0.03[/math] ^ re-arrange to the form [math]x< k[/math]. graph that. Link to comment Share on other sites More sharing options...

daniel_haxby Posted January 31, 2007 Share Posted January 31, 2007 1)3(4d-9)<6(2d-5) for this one ... i end up with 12d-27>12d-30 my question is, where to go from there? Here, you've done everything correctly to this point, except not noticed that you've inexplicably swapped the inequality sign around. If you did that on purpose, you've made a mistake. So go back to check the question. If the question you gave us is correct, then there is no d that satisfies the equation. You can still graph y=x-9/4 and y=x-5/2 to demonstrate graphically that there are no x above the first line which are below the second line. If the question you gave us is incorrect, and the inequality faces the other way, then you got to 12d-27>12d-30. From there you can add 27 to each side, and then you should be able see the answer, and make a graph, although, like last time, you don't need to. 2) |4x-3|=-27 I was told that the two lines (absolute value?, mean two equations) so i ended up 4x-3=-27 and 4x-3=27 I know what is meant by two lines here. A technique taught, which I think is silly, but some people like, is to make two equations which are the equations corresponding to: |x|=x ; for x positive |x|=-x ; for x negative I think that while it is a good method for graphing, it is incredibly confusing unless you understand what absolute value represents, and then why would you need to be told this technique?! The two lines you want are actually: 4x-3=-27 ; for x positive -4x-3=-27 ; for x negative You see, you only change the sign of the x in each case, nothing else. Now you can graph these two lines, using each graph only in its respective domain. An equivalent method is to draw both lines in full, but delete anything falling below the x axis. Wikipedia 'absolute value' to see representative graphs. Now, how do you graph these two lines in particular? The only 'meaningful' way will be to draw x=k, x=-k, delete the bits inappropriate, and then find where x crosses the y axis (!). Link to comment Share on other sites More sharing options...

the tree Posted February 2, 2007 Share Posted February 2, 2007 The two lines you want are actually: 4x-3=-27 ; for x positive -4x-3=-27 ; for x negative No, there is no value of x or definition of f(x) for which |f(x)|=-27, at all. There is nothing to be graphed. Link to comment Share on other sites More sharing options...

daniel_haxby Posted February 2, 2007 Share Posted February 2, 2007 No, there is no value of x or definition of f(x) for which |f(x)|=-27, at all. There is nothing to be graphed. But you can graph the two equations I gave, and point out that no x lies in the image of f(x). Link to comment Share on other sites More sharing options...

daniel_haxby Posted February 2, 2007 Share Posted February 2, 2007 I mean no x=y, hence no solution. Link to comment Share on other sites More sharing options...

the tree Posted February 2, 2007 Share Posted February 2, 2007 I mean no x=y, hence no solution.Huh? Why does there need to be any point in which x=y for any of this?But you can graph the two equations I gave, and point out that no x lies in the image of f(x).Like this? Isn't that a little futile? Link to comment Share on other sites More sharing options...

daniel_haxby Posted February 2, 2007 Share Posted February 2, 2007 I do actually know there is no solution. But if you had to show that graphically, you would graph y=f(x), and look for points y=x. If you do graph the two lines I gave, in their respective domains, then you get a graph that is a horizontal line all above the y axis, so there is no solution. I know it is futile, but the method isn't, just the question. Link to comment Share on other sites More sharing options...

the tree Posted February 2, 2007 Share Posted February 2, 2007 If y=f(x) and [math]f:\to |4x-3|[/math] then there are points where y=x, but that is trivial since you're looking for points where "y=-27", you could plot that but it would be simpler to just state why the question is meaningless, that way you don't have to find a straight edge to draw against. Link to comment Share on other sites More sharing options...

daniel_haxby Posted February 2, 2007 Share Posted February 2, 2007 If you're refusing to graph because you cannot find a straight edge to hand... I know it is trivial. Buuuuut, the guy who posted the questions had obviously only half understood what is a valid technique for solving modulus problems, and while in this case it is unnecessary/pointless, for the future might be worth clarifying. That was all I was trying to explain. but that is trivial since you're looking for points where "y=-27" Yes, I did not explain that bit correctly. I didn't go back and read the posts properly. PS and I wouldn't graph it either, but the original questions were Solve and Graph the following. Link to comment Share on other sites More sharing options...

the tree Posted February 3, 2007 Share Posted February 3, 2007 If you're refusing to graph because you cannot find a straight edge to hand...Yes. I'm a teenager. I often refuse to do most things on a trivial basis. I will not, for instance, go anywhere near a maths room unless I have had, or am currently having or will have when I get there, coffee (I sometimes make an exception when I know there is cake). PS and I wouldn't graph it either, but the original questions were Solve and Graph the following.Either the questions were not copied down correctly, or absolutely no thought was putting into setting them. I would probably have left it as a one dimensional graph if asked to graph it. Link to comment Share on other sites More sharing options...

Khargosh Posted February 7, 2007 Author Share Posted February 7, 2007 Sorry for my rather late reply, havn't had much time to get to a computer. Anyway, the one question, that has been argue'd over, having no sollution, I was told by my teacher, to simply put, No sollution. We've since moved on to thing's like Linear Programing,matricies, and Determinates. I've had a bit of trouble with multiplying matricies, so i can jot down a quick example, and see if you guy's can try to help and see what i'm doing wrong. FIND EACH PRODUCT 1) 2x2 matrix [-3,4] [5 , 2] multiply by 2x2 matrix [1, 0 ] [2, -3] we were told to multiply the first row, from the first matrix, by the first column in the second matrix. so -3x1=-3 and 4x2=8 then second row by second column [this is where i believe i mess up] but i get, 5x0=10 and 2x-3=-6 so the new matrix would then be [-3,8] [10,-6] right after multiplying the first row by the first column, is where i mess up. Any help much appreciated. Link to comment Share on other sites More sharing options...

Bignose Posted February 7, 2007 Share Posted February 7, 2007 You were told by your teacher that multiplying a 2x2 matrix with a 2x2 has no solution? Is that what you are saying, or are you asking for help as to how to correctly multiply these two matrices together? Because what you have done is not right. Each element of the matrix product (the answer, or right hand side) will involve several multiplications and then additions. The 1,1 element of the product is indeed the multiplication of the first matrix's first row by the second matrix's first column. You got the -3 and the 8 correctly. But then you have to add them together to get 5. That is the 1,1 element of the product matrix. The 1,2 element of the product is the multiplication of the first matrix's first row by the second matrix's second column. And then add up the element-by-element results, just like above. The 2,1 element of the product is the multiplication of the first matrix's second row by the second matrix's first column. And then add up the element-by-element results, just like above. The 2,2 element of the product is the multiplication of the first matrix's second row by the second matrix's second column. And then add up the element-by-element results, just like above. I get -6 for the 2,2 element of the product. See if you can't do the other two by yourself now that you know the correct procedure. p.s. 5x0 does not equal 10! Link to comment Share on other sites More sharing options...

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