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X / X-2 - x +1 / x = 8 / x(squared) - 2x

Solve for X

I am lost on how to complete this problem. I need help. I couldnt figure out how to write x squared with the small 2 so I typed it instead. Thanks ! :)

Also I am stumped on this problem as well,

5/ x+6 + 2/x(squared) + 7x + 6 = 3 / x + 1

These are all fractions just incase it appears confusing. I am supposed to write this equation in simplest form. Your help is greatly appreciated !!!!

2007-01-04 06:03:48 · 3 answers · asked by CookFrNW 3 in Science & Mathematics Mathematics

3 answers

You need to use parenthesis, but hopefully I am reading this the way you intended...
x/(x - 2) - (x + 1)/x = 8/(x² - 2x)
Factor the denominator on the last term...
x/(x - 2) - (x + 1)/x = 8/[x(x - 2)]
Multiply through by x(x - 2)...
x² - (x + 1)(x - 2) = 8
Now combine...
x² - x² + x + 2 = 8
x + 2 = 8
x = 6

Check:
x/(x - 2) - (x + 1)/x = 8/(x² - 2x)
6/4 - 7/6 = 8/(36 - 12)
9/6 - 7/6 = 8/24
1/3 = 1/3


Your second problem:
5/(x + 6) + 2/(x² + 7x + 6) = 3/(x + 1)
Factor...
5/(x + 6) + 2/[(x + 6)(x + 1)] = 3/(x + 1)
Multiply by (x + 6)(x + 1)...
5(x + 1) + 2 = 3(x + 6)
Combine and reduce...
5x + 5 + 2 = 3x + 18
2x = 11
x = 11/2
x = 5.5

Check:
5/(x + 6) + 2/(x² + 7x + 6) = 3/(x + 1)
5/11.5 + 2/(30.25 + 38.5 + 6) = 3/6.5
10/23 + 2/74.75 = 6/13
10/23 + 8/299 = 6/13
130/299 + 8/299 = 6/13
138/299 = 6/13
6/13 = 6/13

2007-01-04 06:18:21 · answer #1 · answered by computerguy103 6 · 0 0

Put all of your terms over a common denominator, across both sides. Multiply all terms by x^2 , then multiply all terms by any other denominator term (such as (x-2)). Gather like terms and solve as you would any other equation set. Check your results back in the original equation to make sure your work is right -
good luck!

2007-01-04 06:19:52 · answer #2 · answered by MamaMia © 7 · 0 0

It is unclear as to what is a denominator. Is x - 2 a denominator for the first term?

Try putting brackets around numerators and denominators...it would be very helpful!

2007-01-04 06:19:12 · answer #3 · answered by keely_66 3 · 0 0

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