Difference between revisions of "2002 AMC 10B Problems/Problem 4"
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<math>(10)(17)-(10)(16)+1=170-160+1=11\Longrightarrow\mathrm{ {D} \ }</math> | <math>(10)(17)-(10)(16)+1=170-160+1=11\Longrightarrow\mathrm{ {D} \ }</math> | ||
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| + | One can also do most of the algebra, and only then substitute for <math>x</math>. Note that the first term <math>(3x-2)(4x+1)</math> can be written as | ||
| + | <math>(3x-2)4x + (3x-2)1</math>. Hence we get: | ||
| + | |||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | & (3x-2)(4x+1)-(3x-2)4x+1 \\ | ||
| + | &= (3x-2)4x + (3x-2)1 -(3x-2)4x+1 \\ | ||
| + | &= 3x-2+1 \\ | ||
| + | &= 3x-1 \\ | ||
| + | &= 11. | ||
| + | \end{align*} | ||
| + | </cmath> | ||
==See Also== | ==See Also== | ||
Revision as of 07:46, 2 February 2009
Problem
What is the value of
when 
?
Solution
One can also do most of the algebra, and only then substitute for 
. Note that the first term 
can be written as
. Hence we get:
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
