Difference between revisions of "2010 AMC 10A Problems/Problem 11"
(Created page with 'Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. <math> a\le…') |
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<math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math> | <math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math> | ||
| − | Since we have the range of the solutions, we can make them equal to <math> | + | Since we have the range of the solutions, we can make them equal to <math> 10 </math>. |
<math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math> | <math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math> | ||
Revision as of 17:12, 16 August 2010
Since we are given the range of the solutions, we must re-write the inequalities so that we have 
in terms of 
and 
.
Subtract 
from all of the quantities:
Divide all of the quantities by 
.
Since we have the range of the solutions, we can make them equal to 
.
Multiply both sides by 2.
Simplify
$b-3-a+3 = 20 \RightArrow b-a = 20$ (Error compiling LaTeX. Unknown error_msg)
We need to find 
for the problem, so the answer is 
