Difference between revisions of "1960 AHSME Problems/Problem 28"
Rockmanex3 (talk | contribs) (Solution to Problem 28) |
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==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1960|num-b=27|num-a=29}} | + | {{AHSME 40p box|year=1960|num-b=27|num-a=29}} |
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| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 19:12, 17 May 2018
Problem
The equation 
has:
Solution
Both terms have a 
term, so add 
to both sides. This results in 
.
However, note that if 
is plugged back into the original equation, it results in
![\[3-\frac{7}{0}=3-\frac{7}{0}\]](http://latex.artofproblemsolving.com/a/4/9/a4926d9e1a21982e6b8bd554eb5fad3c2326a6fc.png)
Since dividing by zero is undefined, is an extraneous solution. That means there are no solutions, so the answer is 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
