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| − | == Problem ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 2]] |
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| − | What is the value of
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| − | <math>(3x-2)(4x+1)-(3x-2)4x+1</math>
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| − | when <math>x=4</math>?
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| − | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12 </math>
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| − | == Solution ==
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| − | <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
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| − | <math>(3x-2)4x + (3x-2)1</math>. Hence we get:
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| − | <cmath>
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| − | \begin{align*}
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| − | & (3x-2)(4x+1)-(3x-2)4x+1 \\
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| − | &= (3x-2)4x + (3x-2)1 -(3x-2)4x+1 \\
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| − | &= 3x-2+1 \\
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| − | &= 3x-1 \\
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| − | &= 11.
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| − | \end{align*}
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| − | </cmath>
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=B|num-b=3|num-a=5}}
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| − | [[Category:Introductory Algebra Problems]] | |
