Difference between revisions of "2007 iTest Problems/Problem 2"
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<math>3a + 7b = 1977</math> and <math>5a + b = 2007</math>. | <math>3a + 7b = 1977</math> and <math>5a + b = 2007</math>. | ||
| − | Thus, <math>b=2007-5a</math>, and substituting, <math>3a+14049-35a=1977\Rightarrow -32a=-12072\Rightarrow a=377.25</math>. Thus, <math>b=2007-1886.25\Rightarrow b=120.75</math>. Thus, <math>a+b=377.25+120.75=498\Rightarrow \boxed{\mathrm{B}}</math> | + | Thus, <math>b=2007-5a</math>, and substituting, <math>3a+14049-35a=1977\Rightarrow</math><math>-32a=-12072\Rightarrow a=377.25</math>. Thus, <math>b=2007-1886.25\Rightarrow b=120.75</math>. Thus, <math>a+b=377.25+120.75=498</math><math>\Rightarrow \boxed{\mathrm{B}}</math> |
| + | ==Alternate Solution== | ||
| + | We have <math>3a+7b=1977</math> and <math>5a+b=2007</math>. Notice the symmetry we have if we add the two equations together: <math>8a+8b=3984</math>. Dividing by 8, we have <math>a+b=498</math>. <math>\boxed{\mathrm{B}}</math> | ||
| + | |||
| + | ==See Also== | ||
{{iTest box|year=2007|num-b=1|num-a=3}} | {{iTest box|year=2007|num-b=1|num-a=3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 20:05, 4 February 2023
Problem
Find 
if 
and 
satisfy
and 
.
Solution
and .
Thus, 
, and substituting, 
. Thus, 
. Thus, 
Alternate Solution
We have 
and 
. Notice the symmetry we have if we add the two equations together: 
. Dividing by 8, we have 
. 
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 1 |
Followed by: Problem 3 | |
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