Simple Partial Fractions
The last quiz was an example of partial fractions , i.e. the
technique of decomposing a fraction as a sum of simpler
fractions. This section will consider the simpler forms of this
Find the partial fraction decomposition of 4 / ( x - 4) .
The denominator factorises as x - 4 = ( x - 2)( x +2) . (See the package on
quadratics .) The partial fractions will, therefore, be of the
form a/ ( x - 2) and b/ ( x + 2) . Thus
so that ( a + b ) x + 2( a - b ) = 4
The last line is
( a + b ) x + 2( a - b ) = 4 ,
and this enables a and b to be found. For the equation to be
true for all values of x the coeffcients must match, i.e.
a + b = 0 (coeffcients of x )
2 a - 2 b = 4 (constant terms)
where the first equation holds since there is no x term in 4 /
( x - 4). This set of simultaneous equations may be
solved to give a = 1 and b = - 1. Thus
Find a and b .
Taking common denominators:
so that ( a + b ) x + (2 a - 2 b ) = 4 x. Equating coeffcients
in this case gives
a + b = 4 (coeffcients of x )
2 a - 2 b = 0 (constant terms)
Solving this set of equations gives a = 2 , b = 2. Hence
If which of the following is the solution
to the equation?
Writing all the fractions with a common denominator
so that (3 a + 2 b ) x + (4 a - 3 b ) = x + 7. This gives two
3 a + 2 b = 1 (coeffcients of x )
4 a - 3 b = 7 (constant terms)
Solving these gives a = 1 , b = - 1.