Difference between revisions of "1967 AHSME Problems/Problem 10"
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| + | == Problem == | ||
| + | If <math>\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}</math> is an identity for positive rational values of <math>x</math>, then the value of <math>a-b</math> is: | ||
| + | <math>\textbf{(A)}\ \frac{4}{3} \qquad | ||
| + | \textbf{(B)}\ \frac{5}{3} \qquad | ||
| + | \textbf{(C)}\ 2 \qquad | ||
| + | \textbf{(D)}\ \frac{11}{4} \qquad | ||
| + | \textbf{(E)}\ 3</math> | ||
| + | |||
| + | == Solution == | ||
| + | Given the equation: | ||
| + | <cmath> \frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)} </cmath> | ||
| + | |||
| + | Let's simplify by letting <math>y = 10^x</math>. The equation becomes: | ||
| + | <cmath> \frac{a}{y-1}+\frac{b}{y+2}=\frac{2y+3}{(y-1)(y+2)} </cmath> | ||
| + | |||
| + | Multiplying each term by the common denominator <math>(y-1)(y+2)</math>, we obtain: | ||
| + | <cmath> a(y+2) + b(y-1) = 2y+3 </cmath> | ||
| + | |||
| + | For the equation to be an identity, the coefficients of like terms on both sides must be equal. Equating the coefficients of <math>y</math> and the constant terms, we get the system of equations: | ||
| + | |||
| + | <cmath>\begin{align*} | ||
| + | a+b &= 2 \quad \text{(from coefficients of } y \text{)} \\ | ||
| + | 2a - b &= 3 \quad \text{(from constant terms)} | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | Solving this system, we find: | ||
| + | <cmath> a = \frac{5}{3} \quad \text{and} \quad b = \frac{1}{3} </cmath> | ||
| + | |||
| + | Thus, the difference is: | ||
| + | <cmath> a-b = \boxed{\textbf{(A) } \frac{4}{3}} </cmath> | ||
| + | |||
| + | ~ proloto | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME 40p box|year=1967|num-b=9|num-a=11}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:53, 28 September 2023
Problem
If 
is an identity for positive rational values of 
, then the value of 
is:
Solution
Given the equation:
![\[\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}\]](http://latex.artofproblemsolving.com/3/3/2/33255629cfda7de8b6f83dccb6f4b29ca434abca.png)
Let's simplify by letting 
. The equation becomes:
![\[\frac{a}{y-1}+\frac{b}{y+2}=\frac{2y+3}{(y-1)(y+2)}\]](http://latex.artofproblemsolving.com/c/1/6/c163287ca8588b4b8d69ab9079213ee3056a5e7e.png)
Multiplying each term by the common denominator 
, we obtain:
![\[a(y+2) + b(y-1) = 2y+3\]](http://latex.artofproblemsolving.com/e/8/4/e8416b3cfc32917fff2896fc1b815c95158c9bf6.png)
For the equation to be an identity, the coefficients of like terms on both sides must be equal. Equating the coefficients of 
and the constant terms, we get the system of equations:
Solving this system, we find:
![\[a = \frac{5}{3} \quad \text{and} \quad b = \frac{1}{3}\]](http://latex.artofproblemsolving.com/2/e/c/2ec9a27df6e72687fee50bd5de4d5f21722bff73.png)
Thus, the difference is:
![\[a-b = \boxed{\textbf{(A) } \frac{4}{3}}\]](http://latex.artofproblemsolving.com/2/9/9/29985799ab6b80dcb571a881041bffd6e001bae8.png)
~ proloto
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
