Solving for x (this should be easy but somehow I keep messing it up)

  • Thread starter ilovemynny
  • Start date
In summary, the conversation is about a student starting college and encountering a problem in their vector geometry class. They sought help from others and were able to get their answer cleaned up and clarified.
  • #1
ilovemynny
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Intro:
I'm starting college this fall. I asked one of my professors if we had a textbook and he said that there's an online version and I should read a bit of it if I wanted to.
On the first chapter they were talking about simple (I mean really simple algebra, this is a vector geometry class) but I got completely stumped on one of the problems.

I got an answer but it seems to be different from the textbook answer. I tried putting in number valves for the variable in the 2 different solution but i got different solutions. I probably did this wrong s0 can someone help me solve this I don't want to be stumped on my first week of college with the engineers >.<

Problem:
It should be uploaded on this message as an attachment.

What I Did:

called "problem my way" attachment

Supposed Answer: attachment called "answer"
 

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  • #2
You're answer is the same as the given solution. You just have to clean your expression up a bit.
[tex]x = \frac{e}{\frac{c - a}{a + 1/b} - d} - c[/tex]
Take the fraction in the denominator and multiply by [itex]\frac{b}{b}[/itex]
[tex]x = \frac{e}{\frac{b(c - a)}{ba + 1} - d} - c[/tex]
Perform the subtraction in the denominator by finding the LCD (which is ba + 1):
[tex]x = \frac{e}{\frac{b(c - a)}{ba + 1} - \frac{d(ba + 1)}{(ba + 1)}} - c[/tex]
[tex]x = \frac{e}{\frac{b(c - a) - d(ba + 1)}{ba + 1}} - c[/tex]
Multiply the entire fraction by [itex]\frac{ba + 1}{ba + 1}[/itex]
[tex]x = \frac{e(ba + 1)}{b(c - a) - d(ba + 1)} - c[/tex]


Note to mods: I don't think I'm giving too much away here. This is not in the HW subforum and the OP did show a lot of the work.
 
  • #3
I find it useful to do some substitutions when solving these, you could clean up the expression given and make things a lot easier
 
  • #4
Oh, THANK YOU! I tried using substitution to check but yeah since it's so messy I probably messed up somewhere. Thank you so much! This problem was really starting to freak me out. Thank you, eumayang, for cleaning it up! I tried converting my answer to the book's answer but I was really confused on how to do it.
 
  • #5


I understand that solving problems in mathematics can be challenging, especially when you are just starting college. It is important to remember that making mistakes is a natural part of the learning process and it is important to keep trying and seeking help when needed. It is also important to understand that there can be multiple ways to solve a problem and it is not uncommon for different methods to lead to different solutions.

In this case, it is possible that there was an error in your calculations or that you used a different method than the textbook. I would recommend double-checking your work and trying to solve the problem using the textbook's method to see if you get the same answer. If you are still unsure, don't hesitate to reach out to your professor or classmates for help. It is important to communicate and collaborate with others in your class to improve your understanding and problem-solving skills. Keep practicing and don't be discouraged, you will get the hang of it. Best of luck in your studies!
 

1. How do I know where to start when solving for x?

When solving for x, you should first look for any like terms or constants on either side of the equation. Then, use the inverse operation to isolate the variable on one side of the equation. Once the variable is isolated, solve for x using the appropriate steps.

2. What do I do if there are fractions or decimals in the equation?

To solve for x when there are fractions or decimals involved, you can either multiply both sides of the equation by the common denominator to eliminate the fractions, or you can convert the decimals to fractions by moving the decimal point to the right until it becomes a whole number.

3. Can I solve for x if there are variables on both sides of the equation?

Yes, you can solve for x even if there are variables on both sides of the equation. Use the same steps as you would for isolating a variable on one side, but this time you will need to combine like terms on one side before isolating the variable.

4. How do I check my answer to make sure it is correct?

You can check your answer by plugging it back into the original equation and simplifying both sides. If both sides of the equation are equal, then your answer is correct. You can also graph the equation and see if the x-value you solved for results in the same point on the graph.

5. What should I do if I get a negative value for x?

If you get a negative value for x, it is important to carefully check your work to make sure you did not make a mistake. If you are sure that your work is correct, then the negative value for x is the solution. Keep in mind that some equations may have multiple solutions, so it is important to check all possible solutions.

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