Chikis Posted April 16, 2016 Share Posted April 16, 2016 (edited) I need help with this: I need some explanation when it comes to dealing with partial fractions. [math]\frac{7x+19}{(x+1)(x+5)}[/math] into simpler fractions.[math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math] Why did[math]\frac{7x+19}{(x+1)(x+5)}[/math] become equal to [math]\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math]? Edited April 16, 2016 by Chikis Link to comment Share on other sites More sharing options...
Prometheus Posted April 16, 2016 Share Posted April 16, 2016 (edited) It's not so much that it has become equal but rather that you are supposing it is equal. Having supposed they are equal the task is then to find some A and B which satisfy the equality. It's not necessarily the case that there are solutions for A and B that make the equations equal, in which case the supposition would be mistaken. Edited April 16, 2016 by Prometheus 1 Link to comment Share on other sites More sharing options...
fiveworlds Posted April 16, 2016 Share Posted April 16, 2016 (edited) Well when you are adding two fractions with different denominators you multiply each fraction top and bottom by the denominator of the other fraction so [latex]\frac{a}{b}+\frac{b}{c} = \frac{ac}{bc}+\frac{bb}{bc} = \frac{ac+bc}{bc}[/latex] If you are trying to work the equation back to A+B it isn't really possible. For instance there is two possible values for 7x being [latex]\frac{7-\frac{7}{x+5}}{x+1} [/latex] and [latex]\frac{7-\frac{7}{x+1}}{x+1}[/latex] Edited April 16, 2016 by fiveworlds Link to comment Share on other sites More sharing options...
Chikis Posted April 17, 2016 Author Share Posted April 17, 2016 (edited) Ok, solving [math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math] [math]\rightarrow[/math] [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{1}[/math] How does this give rise to[math]7x+19=A(x+5)+B(x+1)[/math] Edited April 17, 2016 by Chikis Link to comment Share on other sites More sharing options...
swansont Posted April 17, 2016 Share Posted April 17, 2016 x+1 and x+ 5 are common denominators of the fraction. That allows you to easily manipulate the equation You want to rewrite the right-hand side as [math]A(x+5)/(x+1)(x+5) + B(x+1)/(x+1)(x+5)[/math] All you have done is multiply each term by 1 Now the denominator for the whole thing is the same, so it cancels and you are left with the equation in question Link to comment Share on other sites More sharing options...
Prometheus Posted April 17, 2016 Share Posted April 17, 2016 The reason you cannot see how it gives rise to your third line (which is correct) is that your second line has a mistake (check the denominator on the RHS). Correct this mistake and you'll see the equality in the third. Link to comment Share on other sites More sharing options...
Chikis Posted April 17, 2016 Author Share Posted April 17, 2016 [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}[/math] Could this one be right? Is the right hand side correct now? Link to comment Share on other sites More sharing options...
Prometheus Posted April 17, 2016 Share Posted April 17, 2016 Yep. You might be able to just see the equality of the numerators now, or you might want to multiply both sides by the denominator to make it explicit. Link to comment Share on other sites More sharing options...
Chikis Posted April 17, 2016 Author Share Posted April 17, 2016 [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}[/math] Ok! Multiplying the left and right hand side by the denominator, we have [math]A(x+5)+B(x+1)=7x+19[/math] What happens next? Link to comment Share on other sites More sharing options...
Prometheus Posted April 17, 2016 Share Posted April 17, 2016 Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation? Link to comment Share on other sites More sharing options...
swansont Posted April 17, 2016 Share Posted April 17, 2016 Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation? No, you solve for A and B, so that the original equation is true for any x (except where it diverges). You have 2 equations and 2 unknowns. Link to comment Share on other sites More sharing options...
Prometheus Posted April 17, 2016 Share Posted April 17, 2016 Yes, but there are particular values of x (i.e. the roots - of which there are two here) for which it is often easier to see the values of A and B. Link to comment Share on other sites More sharing options...
swansont Posted April 17, 2016 Share Posted April 17, 2016 Yes, but there are particular values of x (i.e. the roots - of which there are two here) for which it is often easier to see the values of A and B. At those values the equation doesn't work. Link to comment Share on other sites More sharing options...
Chikis Posted April 17, 2016 Author Share Posted April 17, 2016 (edited) Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation?Based on the examples I have seen so far, to find A we take that [math]x=-1[/math]So [math]7x-1+19=A(-1+5)+B(-1+1)[/math] [math]\rightarrow[/math] [math]12=4A+0[/math] [math]\rightarrow[/math][math]A=3[/math] To find B, we take that [math]x=-5[/math] So [math]7x-5+19=A(-5+5)+B(-5+1)[/math] [math]\rightarrow[/math] [math]-35+19=0-4B[/math] [math]\rightarrow[/math] [math]-16=-4B[/math] [math]\rightarrow[/math] [math]B=+4[/math] Thus, [math]\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}[/math] Is my work correct? Edited April 18, 2016 by Chikis Link to comment Share on other sites More sharing options...
imatfaal Posted April 19, 2016 Share Posted April 19, 2016 Based on the examples I have seen so far, .... [math]\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}[/math] Is my work correct? Always but always multiply back out to check whether it is correct Link to comment Share on other sites More sharing options...
Chikis Posted April 19, 2016 Author Share Posted April 19, 2016 (edited) Multiplying back? How? Edited April 19, 2016 by Chikis Link to comment Share on other sites More sharing options...
imatfaal Posted April 19, 2016 Share Posted April 19, 2016 Do the reverse - you have just used simultaneous equations to figure out two numbers. Put the numbers back in where your A and B were and check that it comes out correctly Link to comment Share on other sites More sharing options...
Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 (edited) [math]\frac{3}{x+1}+\frac{4}{x+5}[/math] [math]=\frac{3(x+5)+4(x+1)}{(x+1)(x+5)}[/math] [math]=\frac{3x+15+4x+4}{(x+1)(x+5)}[/math] [math]=\frac{7x+19}{(x+1)(x+5)}[/math] If the above is what you meant, I have done it and seen that it came out correctly. Thank you. Edited April 21, 2016 by Chikis 1 Link to comment Share on other sites More sharing options...
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