Difference between revisions of "1967 AHSME Problems/Problem 10"

(Undo Vandalism)
(Solution)
 
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{A}</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 $\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}$ is an identity for positive rational values of $x$, then the value of $a-b$ is:

$\textbf{(A)}\ \frac{4}{3} \qquad \textbf{(B)}\ \frac{5}{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{11}{4} \qquad \textbf{(E)}\ 3$

Solution

Given the equation: \[\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}\]

Let's simplify by letting $y = 10^x$. The equation becomes: \[\frac{a}{y-1}+\frac{b}{y+2}=\frac{2y+3}{(y-1)(y+2)}\]

Multiplying each term by the common denominator $(y-1)(y+2)$, we obtain: \[a(y+2) + b(y-1) = 2y+3\]

For the equation to be an identity, the coefficients of like terms on both sides must be equal. Equating the coefficients of $y$ and the constant terms, we get the system of equations:

\begin{align*} a+b &= 2 \quad \text{(from coefficients of } y \text{)} \\ 2a - b &= 3 \quad \text{(from constant terms)} \end{align*}

Solving this system, we find: \[a = \frac{5}{3} \quad \text{and} \quad b = \frac{1}{3}\]

Thus, the difference is: \[a-b = \boxed{\textbf{(A) } \frac{4}{3}}\]

~ proloto

See also

1967 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png