Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 35"

(Solution)
(Solution)
 
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<cmath>\frac{2}{6-3x}+\frac{5}{x-2}-\frac{3}{4-2x}=\frac{(-2)+6(5)-(-3)}{(2)(3)(x-2)}=\frac{35}{6x-12}</cmath>
 
<cmath>\frac{2}{6-3x}+\frac{5}{x-2}-\frac{3}{4-2x}=\frac{(-2)+6(5)-(-3)}{(2)(3)(x-2)}=\frac{35}{6x-12}</cmath>
  
The answer is A.
+
The answer is <math>\boxed{\text{(B)}}.</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:18, 2 February 2021

Problem

Combine and simplify $\frac{2}{6-3x}+\frac{5}{x-2}-\frac{3}{4-2x}$

$\text{(A)} \quad \frac{-35}{6(x-2)} \quad \text{(B)} \quad \frac{-35}{6(2-x)} \quad \text{(C)} \quad \frac{35}{6(2-x)}$

$\text{(D)} \quad \frac{6}{2-x} \quad \text{(E)} \quad \frac{6}{x-2}$

Solution

The common denominator is $(2)(3)(x-2).$ \[\frac{2}{6-3x}+\frac{5}{x-2}-\frac{3}{4-2x}=\frac{(-2)+6(5)-(-3)}{(2)(3)(x-2)}=\frac{35}{6x-12}\]

The answer is $\boxed{\text{(B)}}.$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
Problem 34
TMTA High School Mathematics Contest Past Problems/Solutions Followed by
Problem 36