Difference between revisions of "1964 AHSME Problems/Problem 11"
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<math>0 = y - 6</math> | <math>0 = y - 6</math> | ||
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+ | <math>y = 6</math> | ||
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+ | Plugging that back in to <math>x = 3y + 3</math> gives <math>x = 3(6) + 3</math>, or <math>x = 21</math>. Thus, <math>x+y = 21 + 6</math>, or <math>x+y=27</math>, which is option <math>\boxed{\textbf{(D)}}</math> | ||
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==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 11:33, 23 July 2019
Problem
Given and , find the value of
Solution
Since and , we have:
and
Note that if , then . Setting the exponents equal gives and . Plugging the first equation into the second equation gives:
Plugging that back in to gives , or . Thus, , or , which is option
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
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