Difference between revisions of "1955 AHSME Problems/Problem 47"

(Solution)
(Solution)
 
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<cmath>a = a^2 + ab + ac</cmath>
 
<cmath>a = a^2 + ab + ac</cmath>
 
<cmath>a((a + b + c) - 1) = 0</cmath>
 
<cmath>a((a + b + c) - 1) = 0</cmath>
These expressions are equal only when <math>\boxed{\text{(C) equal whenever }  a + b + c = 1.}</math>
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These expressions are <math>\boxed{\text{(C) equal whenever }  a + b + c = 1.}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:54, 24 July 2022

Problem 47

The expressions $a+bc$ and $(a+b)(a+c)$ are:

$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$

Solution

Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \[a + bc = a^2 + ab + ac + bc\] \[a = a^2 + ab + ac\] \[a((a + b + c) - 1) = 0\] These expressions are $\boxed{\text{(C) equal whenever }  a + b + c = 1.}$

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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All AHSME Problems and Solutions


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