Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1950 AHSME Problems/Problem 6"

m (→‎Solution)
 
(5 intermediate revisions by 3 users not shown)
Line 8: Line 8:
 
If we solve the second equation for <math>x</math> in terms of <math>y</math>, we find <math>x=-\dfrac{y+3}{2}</math> which we can substitute to find:
 
If we solve the second equation for <math>x</math> in terms of <math>y</math>, we find <math>x=-\dfrac{y+3}{2}</math> which we can substitute to find:
  
−
<math>2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0</math>
+
<cmath>2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0</cmath>
  
−
Multiplying by four and simplifying, we find:
+
Multiplying by two and simplifying, we find:
  
−
<math>2*(2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1)=2*0</math>
+
<cmath>\begin{align*}
−
<math>(y+3)^2 -6y-18+10y+2=0</math>
+
2\cdot[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1]&=2\cdot 0\\
−
<math>y^2+6y+9-6y-18+10y+2=0</math>
+
(y+3)^2 -6y-18+10y+2&=0\\
−
<math>y^2+10y-7=0</math>
+
y^2+6y+9-6y-18+10y+2&=0\\
 +
y^2+10y-7&=0
 +
\end{align*}</cmath>
  
−
This is answer <math>\textbf{(C)}</math> so the answer is <math>\boxed{\textbf{(C)}\ y^{2}+10y-7=0}</math>
+
Therefore 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 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 $y$ which will satisfy the equations $2x^{2}+6x+5y+1=0, 2x+y+3=0$ may be found by solving:

$\textbf{(A)}\ y^{2}+14y-7=0\qquad\textbf{(B)}\ y^{2}+8y+1=0\qquad\textbf{(C)}\ y^{2}+10y-7=0\qquad\\ \textbf{(D)}\ y^{2}+y-12=0\qquad \textbf{(E)}\ \text{None of these equations}$

Solution

If we solve the second equation for $x$ in terms of $y$, we find $x=-\dfrac{y+3}{2}$ which we can substitute to find:

\[2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0\]

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*}

Therefore the answer is $\boxed{\textbf{(C)}\ y^{2}+10y-7=0}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_6&oldid=56122"