# Algebraic fractions: Question e: 5/[x (x + 1)] - 2/x + 3/(x + 1)

• 02-08-2018, 01:32 AM
Algebraic fractions: Question e: 5/[x (x + 1)] - 2/x + 3/(x + 1)
Question:

. . . . .[tex]\mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, -\, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}[/tex]

I'm getting myself confused with this one.

1) I multiplied all the denominators, so they all became x(x+1)x(x+1).
2) Then 5x(x+1) - 2x(x+1)(x+1) + 3x(x+1)x
3) Becomes, 5x2 + 5x - 2x3 - 4x2 - 2x + 3x3 + 3x2
4) Simplified, 4x2 + 3x + x3 / x(x+1)x(x+1).

The answer is x + 3 / x(x+1). I don't know how to get this.

Thxs.
• 02-08-2018, 03:07 AM
mmm4444bot
Quote:

Question:

. . . . .[tex]\mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, -\, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}[/tex]

I'm getting myself confused with this one.

… Simplified, [4x2 + 3x + x3]/[x(x+1)x(x+1)]

The answer is [x + 3]/[x(x+1)]

I don't know how to get this.

Notice that x^3 + 4x^2 + 3x can be factored (to start, every term contains at least one factor of x). If you completely factor this numerator, you can then cancel common factors, to get the known answer.

Also, it's true that x(x+1)x(x+1) is a common denominator, but it's not the easiest one to use.

In the given expression, the three denominators are x(x+1), x, and x+1.

Can you see that x(x+1) is also a common denominator? :idea: Using that one (at the beginning) is less work.
• 02-08-2018, 03:56 AM
Quote:

Originally Posted by mmm4444bot
Notice that x^3 + 4x^2 + 3x can be factored (to start, every term contains at least one factor of x). If you completely factor this numerator, you can then cancel common factors, to get the known answer.

Also, it's true that x(x+1)x(x+1) is a common denominator, but it's not the easiest one to use.

In the given expression, the three denominators are x(x+1), x, and x+1.

Can you see that x(x+1) is also a common denominator? :idea: Using that one (at the beginning) is less work.

Thanks mmm4444bot. So much easier with x(x+1). Don't know why I didn't see that.

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

5 - 2x - 2 + 3x = x+3

Thus, x+3/x(x+1)

Many thanks.
• 02-08-2018, 10:24 AM
Denis
Quote:

Thus, x+3/x(x+1)

On a test, that would be considered incorrect; more bracketing required:
[x+3] / [x(x+1)]
• 02-08-2018, 03:10 PM
mmm4444bot
Quote:

5/[x(x+1)] - 2(x+1)/[x(x+1)] + 3x/[x(x+1)]

5 - 2x - 2 + 3x = x+3

Thus, [x+3]/[x(x+1)]

As Denis noted, we need to put grouping symbols around some numerators and denominators, when typing algebraic ratios with a keyboard. This is to ensure that people reading these texted expressions know what's on top and what's on bottom. Let us know, if you're not sure about this. :cool:
• 02-08-2018, 10:42 PM