Question 169958
the sum 2/x+1 + 7/x+2 can be expressed in the form ax+b/(x+1)(x+2). what are the values for a and b?
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There are two ways to do this:

First way, by substitution of x=0 and x=1.

{{{2/(x+1) + 7/(x+2) = (ax+b)/(x+1)(x+2)}}}

Substitute x=0:

{{{2/(x+1) + 7/(x+2) = (ax+b)/((x+1)(x+2))}}}

{{{2/(0+1) + 7/(0+2) = (a*0+b)/((0+1)(0+2))}}}

{{{2/1 + 7/2 = b/((1)(2))}}}
{{{2 + 7/2 = b/2}}}
Multiply through by 2
{{{4 + 7 = b}}}
{{{11=b}}}

Substitute {{{ll}}} for {{{b}}}:

{{{2/(x+1) + 7/(x+2) = (ax+11)/(x+1)(x+2)}}}

Substitute x=1:

{{{2/(x+1) + 7/(x+2) = (ax+11)/((x+1)(x+2))}}}

{{{2/(1+1) + 7/(1+2) = (a*1+11)/((1+1)(1+2))}}}

{{{2/2 + 7/3 = (a+11)/((2)(3))}}}
{{{1 + 7/3 = (a+11)/6}}}
Multiply through by 6
{{{6 + 14 = a+11}}}
{{{20=a+11}}}
{{{9=a}}}

So a=9 and b=11

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Second way:

{{{2/(x+1) + 7/(x+2)}}}

Get LCD=(x+1)(x+2)

{{{(2(x+2))/(x+1)(x+2) + (7(x+1))/(x+2)(x+1)}}}

{{{(2x+4)/(x+1)(x+2) + (7x+7)/(x+2)(x+1)}}}

{{{(2x+4+7x+7)/(x+1)(x+2)}}}

{{{(9x+11)/(x+1)(x+2)}}}

Compare that with

{{{(ax+b)/(x+1)(x+2)}}}

and obviously {{{a=9}}} and {{{b=11}}}

Edwin</pre>