Solve Algebra Word Problem: x+1/x - 1/4 = x+3/x+2

AI Thread Summary
To solve the equation x + 1/x - 1/4 = x + 3/x + 2, the first step is to find a common denominator, which is 4x(x+2). The equation can be rewritten and simplified by combining fractions on both sides. After simplifying, the equation leads to -6x/4x(x+2) = 0, resulting in the solution x = 0. This value can be verified by substituting it back into the original equation to ensure it satisfies the condition.
homegrown898
Messages
16
Reaction score
0
http://www26.brinkster.com/nick898/question.htm

I`ve scanned the question so you can understand it better. This is how
I started...

x+1/x - 1/4 = x+3/x+2

But this is how far I got. I don`t know what do after this part and I
don`t even know if what I`m doing is correct.
 
Last edited by a moderator:
Physics news on Phys.org
I'll show what I got..
The original one should be (x+1)/x
And it's saying that (x+3)/(x+2) = (x+1)/x - 1/4
So you aren't far off... not you can use your favourite method of solving rational expressions.. my way of choice is cross-multiply...
In the end, x = -4 || 2
Then go on from there..
 
Can anyone help me?

To solve this algebra word problem, we first need to find a common denominator for all the fractions involved. In this case, the common denominator is 4x(x+2). We can rewrite the equation as:

4x(x+2)(x+1)/4x(x+2) - (x+2)/4x(x+2) = 4x(x+2)(x+3)/4x(x+2) + (x+1)/4x(x+2)

Now, we can combine the fractions on the left side and the fractions on the right side:

(4x^2 + 8x + 4 - x - 2)/4x(x+2) = (4x^2 + 12x + 6 + x + 1)/4x(x+2)

Simplifying further, we get:

(4x^2 + 7x + 2)/4x(x+2) = (4x^2 + 13x + 7)/4x(x+2)

Now, we can subtract both sides by (4x^2 + 13x + 7)/4x(x+2):

(4x^2 + 7x + 2)/4x(x+2) - (4x^2 + 13x + 7)/4x(x+2) = 0

Simplifying, we get:

-6x/4x(x+2) = 0

Dividing both sides by -6/4x(x+2), we get:

x = 0

Therefore, the solution to this algebra word problem is x = 0. You can plug in this value for x in the original equation to check if it satisfies the equation. I hope this helps!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top