Difference between revisions of "1985 AHSME Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
From <math>2x = 7</math> (as above), we can directly compute <cmath>\begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*}</cmath> so <math>4x+1 = 14+1 = \boxed{\text{(A)} \ 15}</math>. | From <math>2x = 7</math> (as above), we can directly compute <cmath>\begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*}</cmath> so <math>4x+1 = 14+1 = \boxed{\text{(A)} \ 15}</math>. | ||
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| + | ==Solution 3== | ||
| + | Multiply both sides of the equation by <math>2</math> to get <math>4x + 2 = 16</math>. Then subtract <math>1</math> from both sides to get <math>4x + 1 = \boxed{\text{(A)} \ 15}</math> | ||
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| + | -j314andrews | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|before=First Problem|num-a=2}} | {{AHSME box|year=1985|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:08, 3 July 2025
Problem
If 
, then 
Solution 1
We have 
so 
Solution 2
From 
(as above), we can directly compute 
so 
.
Solution 3
Multiply both sides of the equation by 
to get 
. Then subtract 
from both sides to get 
-j314andrews
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
