Answer:

Exercise 1.26

1.8 Simplification of fractions (EMAQ)

We have studied procedures for working with fractions in earlier grades.

ab×cd=acbd(b≠0;d≠0)ab×cd=acbd(b≠0;d≠0)

ab+cb=a+cb(b≠0)ab+cb=a+cb(b≠0)

ab÷cd=ab×dc=adbc(b≠0;c≠0;d≠0)ab÷cd=ab×dc=adbc(b≠0;c≠0;d≠0)

Note: dividing by a fraction is the same as multiplying by the reciprocal of the fraction.

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

x2+3xx+3

x2+3xx+3

has a quadratic binomial in the numerator and a linear binomial in the denominator. We have to apply the different factorisation methods in order to factorise the numerator and the denominator before we can simplify the expression.

x2+3xx+3=x(x+3)x+3=x(x≠−3)

x2+3xx+3=x(x+3)x+3=x(x≠−3)

If x=−3x=−3 then the denominator, x+3=0x+3=0 and the fraction is undefined.