Difference between revisions of "1950 AHSME Problems/Problem 14"

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== Problem==
 
== Problem==
  
For the simultaneous equations <cmath>2x-3y=8\\6y-4x=9</cmath>
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For the simultaneous equations <cmath>2x-3y=8</cmath>
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<cmath>6y-4x=9</cmath>
  
 
<math> \textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions} </math>
 
<math> \textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions} </math>
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Something is clearly contradictory so <math>\boxed{\mathrm{(D)}\text{ There is no solution}.}</math>
 
Something is clearly contradictory so <math>\boxed{\mathrm{(D)}\text{ There is no solution}.}</math>
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Alternatively, note that the second equation is a multiple of the first except that the constants don't match up. So, there is no solution.
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:15, 21 December 2015

Problem

For the simultaneous equations \[2x-3y=8\] \[6y-4x=9\]

$\textbf{(A)}\ x=4,y=0\qquad\textbf{(B)}\ x=0,y=\frac{3}{2}\qquad\textbf{(C)}\ x=0,y=0\qquad\\ \textbf{(D)}\ \text{There is no solution}\qquad\textbf{(E)}\ \text{There are an infinite number of solutions}$

Solution

Try to solve this system of equations using the elimination method.

\begin{align*} -2(2x-3y)&=-2(8)\\ 6y-4x&=-16\\ 6y-4x&=9\\ 0&=-7 \end{align*}

Something is clearly contradictory so $\boxed{\mathrm{(D)}\text{ There is no solution}.}$

Alternatively, note that the second equation is a multiple of the first except that the constants don't match up. So, there is no solution.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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