1. ## Help with Algebra??

This question is off topic. I have to take a Placement Test for college to see what classes that should put me in. So i got a library book on forgotten algebra and i need some help. I will do the best i can to write out these problems because i do not have a fancy algebra, trig and calculus and so on keyboard. Here are the problems as follow. Before i start. The are a few things. I need to know how you got the answer. Has anyone did math long distance over the internet before??

5-2x
------- = -2
x-1

2
-- + 3(x-1)
x -------- =1
5x

4
---- - 1
x-2 ---- =5
x ------
x-2

&quot;There are 3 separate problems. I think you have to cross mutiply. I had to use the hyphen mark so it&#039;s all mess up and side ways. Please reply back. Thanks for any help.&quot;

2. ## Re:Help with Algebra??

These seem to be very simple problems, however it is very hard to figure out how the problem is layed out, especially on the last two. I know it is very hard to do using simple characters. Can you write them out lined up a little better?

The first one should go like this:

5-2x
------- = -2
x-1

Multiply both sides by (x-1):
5-2x = -2(x-1)
Let&#039;s break up the paranthesis on the right hand side:
5-2x = -2x - (-2)
5-2x = -2x + 2
Let&#039;s add 2x to both sides:
5 = 2

It has no solution ...

3. ## Re:Help with Algebra??

Do you have a scan of these problems instead - I highly doubt ACSII formated by the forum is a good solution.

anyways Aleabra is very simple, the approach cloverm showed you will solve any algebra problem on that level.

4. ## Re:Help with Algebra??

These seem to be very simple problems, however it is very hard to figure out how the problem is layed out, especially on the last two. I know it is very hard to do using simple characters. Can you write them out lined up a little better?

The first one should go like this:

5-2x
------- = -2
x-1

Multiply both sides by (x-1):
5-2x = -2(x-1)
Let&#039;s break up the paranthesis on the right hand side:
5-2x = -2x - (-2)
5-2x = -2x + 2
Let&#039;s add 2x to both sides:
5 = 2

It has no solution ...

[/quote]

Ok I may have done this wrong, but I may have to disagree with you cloverm. I agree up to this part:
5-2x = 2x-2

But I would do this from here on out:
5 - 2x +2 = 2x - 2 +2
7-2x=2x
7-2x+2x=2x+2x
7=4x
7/4=x

Like I said this has a very high likely hood of being wrong, its been a LONG TIME.

&lt;edit&gt;
Ok I was very wrong! :-[ I checked my solution, it doesn&#039;t work out! You are the master cloverm!! ;D
&lt;/edit&gt;

5. ## Re:Help with Algebra??

Ok I may have done this wrong, but I may have to disagree with you cloverm. I agree up to this part:
5-2x = 2x-2

[/quote]
No, no, you left out the - sign on the right hand side:

5-2x = -2(x-1)

6. ## Re:Help with Algebra??

Ah yes that would change things quite a bit. Like I said I&#039;m just dumb! :-[ :-[ :-[ My appologies!

7. ## Re:Help with Algebra??

Thanks guys. Here are the other 2 problems. I will do my best to line them up. I have not had algebra in about 2 years. The last algebra i had was Algebra 1.

2 3(x-1)
--- + -------- =1
x 5x

4 1
----- - ----- = 5
x-2 x ------
x-2

8. ## Re:Help with Algebra??

If I&#039;m correct:

1)

Contraction of division:
10x/5x^2 + (3(x-1)x)/5x^2 = 1
(10x + 3(x-1)x) / 5x^2 = 1

Isolate x
(3x^2 + 7x) / 5x^2 =1
3x^2 + 7x = 5x^2
3x + 7 = 5x
-2x = -7

x = (7/2)

control:
2/(7/2) + (3((7/2) - 1)) / 5(7/2)) = 1

~0.57 + ~0.43 = 1

This seems correct!

2)
Same as before:

Contraction:
4/(x-2) - 1/x = 5/(x-2)

4/(x-2) - 5/(x-2) = 1/x

-1 / (x-2) = 1/x

-1 = (x-2) / x
-1 = 1 - 2/x
-2 = -2/x
-2x = -2
x = 1

control:

4 / ( -1) - (1 / 1) = 5 / ( -1)
-4 -1 = -5
-4 = - 4

this also seems correct

-edit-

oh and damn, you made me do math during my vaction.. I had to think... that hurt.

9. ## Re:Help with Algebra??

Lovechild. How the hell did you do that?? Now my head hurts because you have me thinking. Is there an easier way around these. I have this feeling when i take the test they are going to place me in arithmetic because i had not take any algebra in about 2 years. I&#039;m seriously going brain dead. Please show me an easier way. Please reply back. Thanks Lovechild.

10. ## Re:Help with Algebra??

Okay basically the first step is always to contract the multiple divisions into one common division (by getting a common divisor)

2/x + 1/y would then be (2*y)/(x*y) + (1*x)/(y*x)
and contracted it would be (2y + x) / xy

This way you can proced pretty much like you would always do, this operation is legal because you infact only extend the division by 1 (x/x is always 1, as is y/y)

Always extend the one division by the others divisor and vice virsa to get a common divisor . and thus be able to get only one division to work with.

now is 2/x + 1/y = 2
then

(2y + x) / xy = 2
2y +x = 2xy

and so and and so forth (disgard the fact that is has infinte outcomes due to the fact that you have multiple variables - it&#039;s just to show how it&#039;s done).

the rest is simply to isolate the variable (if you have more variables this calls for some more magic but that&#039;s no harder you basically need as many different equations as you have variables to solve that).

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