Difference between revisions of "1967 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
− | <math>\ | + | 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 == | == See also == |
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:
Let's simplify by letting . The equation becomes:
Multiplying each term by the common denominator , we obtain:
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:
Thus, the difference is:
~ 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 |
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