Difference between revisions of "1950 AHSME Problems/Problem 6"
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==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1950|num-b=5|num-a=7}} | + | {{AHSME 50p box|year=1950|num-b=5|num-a=7}} |
| − | [[ | + | [[Category:Introductory Algebra Problems]] |
| + | {{MAA Notice}} | ||
Latest revision as of 10:57, 5 July 2013
Problem
The values of 
which will satisfy the equations 
may be found by solving:
Solution
If we solve the second equation for 
in terms of , we find 
which we can substitute to find:
![\[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0\]](http://latex.artofproblemsolving.com/a/c/8/ac834f6e09eba6abc42d3d4beae73746f9b0a190.png)
Multiplying by two and simplifying, we find:
![\begin{align*} 2\cdot[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1]&=2\cdot 0\\ (y+3)^2 -6y-18+10y+2&=0\\ y^2+6y+9-6y-18+10y+2&=0\\ y^2+10y-7&=0 \end{align*}](http://latex.artofproblemsolving.com/1/e/d/1ed85234cce2f3e8d4e9e58231335662a760caee.png)
Therefore the answer is 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
