Prove that...?

Question

Prove that:
(x + 2)/x - (x-1)/(x+1) = 2(2x + 1)/x(x+1)

Method

Lol im not bad at maths. Not the greatest either. Just struggle on harder questions...

Answer 1

The lowest common denominator is x(x+1)

So, multiply the first term (x+2)/x by (x+1)/(x+1) This results in (x+2)(x+1)/x(x+1).

Now, multiply the second term (x-1)/(x+1) by x/x. This results in x(x-1)/x(x+1)

So adding together you get:

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

Now, multiply out the terms = (x^2+3x+2-x^2+x)/x(x+1)

That simplifies to: 4x+2/x(x+1)

Now, factor out a 2 in the numerator, and you get:

2(2x+1)/x(x+1) which is what you wanted.

Answer 2

Rewrite the LHS as a single fraction, using the LCM: x(x + 1).
[(x + 2)(x + 1) - x(x - 1)] / [x(x + 1)] =
[x^2 + 3x + 2 - x^2 + x] / [x(x + 1)] =
[4x + 2] / [x(x + 1)] =
[2(2x + 1)] / [x(x + 1)].

Answer 3

OK, you need to make both parts of the left side of the equal sign be over the same denominator: x(x+1)

(x+2)/x = (x+1)(x+2)/x(x+1) Agreed?

(x-1)/(x+1) = x(x-1)/x(x+1) Agreed?

And so on...

Answer 4

LHS
[ (x + 2)(x + 1) - x.(x - 1) ] / [ x.(x + 1) ]
(x² + 3x + 2 - x² + x) / [ x.(x + 1) ]
(4x + 2) / [ x.(x + 1) ]
2.(2x + 1) / [ x.(x + 1) ] = RHS

Answer 5

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

=([x+2)(x+1) - x(x-1)]/x.(x+1)

= (x^2+3x+2-x^2+x)/x.(x+1)
= (4x+2)/x.(x+1)
=2(2x+1)/x(x+1) q.e.d

Answer 6

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