Difference between revisions of "1956 AHSME Problems/Problem 17"
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| + | == Problem 17== | ||
| + | |||
| + | The fraction <math>\frac {5x - 11}{2x^2 + x - 6}</math> was obtained by adding the two fractions <math>\frac {A}{x + 2}</math> and | ||
| + | <math>\frac {B}{2x - 3}</math>. The values of <math>A</math> and <math>B</math> must be, respectively: | ||
| + | |||
| + | <math> \textbf{(A)}\ 5x,-11\qquad\textbf{(B)}\ -11,5x\qquad\textbf{(C)}\ -1,3\qquad\textbf{(D)}\ 3,-1\qquad\textbf{(E)}\ 5,-11 </math> | ||
| + | |||
This is essentially asking for the partial fraction decomposition of <cmath>\frac{5x-11}{2x^2 + x - 6}</cmath> | This is essentially asking for the partial fraction decomposition of <cmath>\frac{5x-11}{2x^2 + x - 6}</cmath> | ||
Looking at <math>A</math> and <math>B</math>, we can write | Looking at <math>A</math> and <math>B</math>, we can write | ||
| − | < | + | <cmath>A(2x-3) + B(x+2) = 5x-11</cmath>. |
| − | Substituting < | + | Substituting <math>x= -2</math>, we get |
<cmath>A(-4-3) + B(0) = -21 \Rrightarrow A = 3</cmath> | <cmath>A(-4-3) + B(0) = -21 \Rrightarrow A = 3</cmath> | ||
| − | Substituting < | + | Substituting <math>x = \frac{3}{2}</math>, we get |
<cmath>A(0) + B(\frac{7}{2}) = \frac{15}{2} - \frac{22}{2} \Rrightarrow B = -1</cmath> | <cmath>A(0) + B(\frac{7}{2}) = \frac{15}{2} - \frac{22}{2} \Rrightarrow B = -1</cmath> | ||
| − | Thus, our answer is < | + | Thus, our answer is <math>\boxed{D}</math> |
~JustinLee2017 | ~JustinLee2017 | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME box|year=1956|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:36, 12 February 2021
Problem 17
The fraction 
was obtained by adding the two fractions 
and
. The values of 
and 
must be, respectively:
This is essentially asking for the partial fraction decomposition of ![\[\frac{5x-11}{2x^2 + x - 6}\]](http://latex.artofproblemsolving.com/3/2/4/324a4d49a88c01b257239e396b0599a533f6028d.png)
Looking at and , we can write
![\[A(2x-3) + B(x+2) = 5x-11\]](http://latex.artofproblemsolving.com/0/8/6/086d190caa252e594851305113d5bd5aa3d9347d.png)
.
Substituting 
, we get
![\[A(-4-3) + B(0) = -21 \Rrightarrow A = 3\]](http://latex.artofproblemsolving.com/1/b/b/1bbada5205ede167ec67a5bc01a5531bee9a6892.png)
Substituting 
, we get
![\[A(0) + B(\frac{7}{2}) = \frac{15}{2} - \frac{22}{2} \Rrightarrow B = -1\]](http://latex.artofproblemsolving.com/4/4/0/440c4986ccc4180c3d2f71702eec4b1de600ff20.png)
Thus, our answer is 
~JustinLee2017
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
