Hard Problem.....

[michiroadie]

Hello all....I don't really get this problem....

5/2 - 3/2x = x

solve for all xs'

 

[Mrspi]

Hello all....I don't really get this problem....

5/2 - 3/2x = x

solve for all xs'

I think you need to clarify the problem.

Do you mean this?

(5/2) - (3/2)*x = x

Or this?

(5/2) - 3/(2x) = x

Please let us know which you mean, because it DOES make a difference.

 

[michiroadie]

Sorry, I meant

(5/2) - (3/2x) = x

I did do the problem, however I got one answer which was 1, but the book listed the answer as (3/2) as well.

 

[Mrspi]

Sorry, I meant

(5/2) - (3/2x) = x

I did do the problem, however I got one answer which was 1, but the book listed the answer as (3/2) as well.

Thanks for the clarification. As long as x is NOT equal to 0, you can multiply both sides of the equation by 2x, which is the common denominator for the fractions:

2x*(5/2) - 2x*(3/2x) = 2x*x

5x - 3 = 2x^2

Is that the equation you worked with? Show us what you did to solve it....and perhaps we can spot a mistake.

 

[michiroadie]

Sorry, I meant

(5/2) - (3/2x) = x

I did do the problem, however I got one answer which was 1, but the book listed the answer as (3/2) as well.

Thanks for the clarification. As long as x is NOT equal to 0, you can multiply both sides of the equation by 2x, which is the common denominator for the fractions:

2x*(5/2) - 2x*(3/2x) = 2x*x

5x - 3 = 2x^2

Is that the equation you worked with? Show us what you did to solve it....and perhaps we can spot a mistake.

This is what I had done:

5/2 (x2) - 3/2x (x2) = x (x2)
5 - 3x = 2x
5 = 3x + 2x
5 = 5x
5/5 = 5/5x
1 = x

 

[Mrspi]

This is what I had done:

5/2 (x2) - 3/2x (x2) = x (x2)
5 - 3x = 2x
5 = 3x + 2x
5 = 5x
5/5 = 5/5x
1 = x

Hmmm...what do you mean by x2??

One fraction has a denominator of 2. The other fraction has a denominator of 2x. The least common multiple of 2 and 2x is just 2x.

2x*(5/2) - 2x*(3/2x) = 2x*x

The 2's divide out on the first term. the "2x"s divide out on the second term. And 2x*x is 2x[sup:295uaowe]2[/sup:295uaowe]

So you're left with this:

x*5 - 3 = 2x[sup:295uaowe]2[/sup:295uaowe]

or,

5x - 3 = 2x[sup:295uaowe]2[/sup:295uaowe]

I suggest that you get one side equal to 0, and then either use the quadratic formula, or try to factor the non-zero side.

 

[michiroadie]

This is what I had done:

5/2 (x2) - 3/2x (x2) = x (x2)
5 - 3x = 2x
5 = 3x + 2x
5 = 5x
5/5 = 5/5x
1 = x

Hmmm...what do you mean by x2??

One fraction has a denominator of 2. The other fraction has a denominator of 2x. The least common multiple of 2 and 2x is just 2x.

2x*(5/2) - 2x*(3/2x) = 2x*x

The 2's divide out on the first term. the "2x"s divide out on the second term. And 2x*x is 2x[sup:243o0wr7]2[/sup:243o0wr7]

So you're left with this:

x*5 - 3 = 2x[sup:243o0wr7]2[/sup:243o0wr7]

or,

5x - 3 = 2x[sup:243o0wr7]2[/sup:243o0wr7]

I suggest that you get one side equal to 0, and then either use the quadratic formula, or try to factor the non-zero side.

oh ok thanks!