Maths Boffs...Help!

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m_2005

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I had some work set for Maths, and I just need to check if my answers were right, so Insha'allah could someone post the answers, so I can compare. The topic is simplifying Algebraic Fractions

Simplify

1 a) 5 + 3/(x+2) =
b) 4/(3-2x) - 1/(1+x) =
c) x/x^2-x-6 - (x+1)/x^2-9 =
d) (x+4)/7 * 14/x^2-16 =
e) (2x+4)/(x-5) ÷ (x+2)/x^2 - 4x-5 =

Note:

Just to make things clear

* means multiply
( ) The bracks are there just to show things are grouped
/ This is the line seperating Numerator and Denominator


Jazakallah in Advance!
 
:sl:

Here are my answers:

1. (a). (5x+13)/(x+2)

(b). (1+6x)/(3-2x)(1+x)

(c). (-2)/(x+2)(x+3)(x-3)

(d). 2/(x-4)

(e). 2(x+1)

:w:
 
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Assalamu alaikum.

My answers are below:

a) (5x+13)/(x+2)

Just multiply the 5 by (x+2) and divide by (x+2), then you will have a common denominator which means you can add these fractions. Add and simplify.

b) (1+6x)/(1+6)(3-2x)

Do the same. Multiply the 4 by (1+x) and divide by the same. Multiply the -1 by (3-2x) and divide by the same. Now you have a fraction with a common denominator. Add and simplify, paying special attention to signs when expanding -(3-2x) .

c) -2/(x+3)(x-3)(x+2)

Do the same as b). Get a common denominator. Add the fraction. Do not alter the denominator as it is already fully factorised, which means it is already in its simplest form. Expand the numerator, paying special attention to signs when expanding -(x+1)(x+2).

d) 2/(x-4)

First factorise the denominator of the right hand fraction x^2-16 using 'difference of two squares'. Cancel (x+4) from the numerator and the denominator and simplify 14/7 to 2.

e) 2(x+1)

Factorise the numerator of the 'numerator fraction'. Factorise the denominator of the 'denominator fraction'. When two fractions divide, you can simply switch the numerator and the denominator around of the the dividing fraction and multiply the two fractions as normal. Cancel common factors from the numerator and the denominator to leave you with the answer 2(x+1).
 
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