Difference between revisions of "1950 AHSME Problems/Problem 6"
(Created page with "== Problem== The values of <math>y</math> which will satisfy the equations <math> 2x^{2}+6x+5y+1=0, 2x+y+3=0 </math> may be found by solving: <math> \textbf{(A)}\ y^{2}+14y-7=0...") |
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| Line 17: | Line 17: | ||
<math>y^2+10y-7=0</math> | <math>y^2+10y-7=0</math> | ||
| − | This is answer <math>\textbf{C}</math> so the answer is <math>\textbf{(C)}\ y^{2}+10y-7=0</math> | + | This is answer <math>\textbf{(C)}</math> so the answer is <math>\boxed{\textbf{(C)}\ y^{2}+10y-7=0}</math> |
==See Also== | ==See Also== | ||
{{AHSME box|year=1950|num-b=5|num-a=7}} | {{AHSME box|year=1950|num-b=5|num-a=7}} | ||
Revision as of 14:08, 28 October 2011
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:
Multiplying by four and simplifying, we find:
This is answer 
so the answer is 
See Also
| 1950 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
