Difference between revisions of "1975 AHSME Problems/Problem 2"

m (Solution)
 
Line 1: Line 1:
 +
==Problem==
 +
 
For which real values of m are the simultaneous equations  
 
For which real values of m are the simultaneous equations  
<cmath>  
+
 
\begin{align*}y &= mx + 3 \\  y& = (2m - 1)x + 4\end{align*} </cmath>  
+
<cmath> \begin{align*}y &= mx + 3 \\  y& = (2m - 1)x + 4\end{align*} </cmath>  
 +
 
 
satisfied by at least one pair of real numbers <math>(x,y)</math>?  
 
satisfied by at least one pair of real numbers <math>(x,y)</math>?  
  
<math>\textbf{(A)}\ \text{all }m\qquad
+
<math>\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\  
\textbf{(B)}\ \text{all }m\neq 0\qquad
+
\text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m </math>   
\textbf{(C)}\ \text{all }m\neq 1/2\qquad
 
\textbf{(D)}\ \text{all }m\neq 1\qquad \\
 
\textbf{(E)}\ \text{no values of }m </math>   
 
  
  
Line 17: Line 17:
 
Solving the systems of equations, we find that <math>mx+3 = (2m-1)x+4</math>, which simplifies to <math>(m-1)x+1 = 0</math>. Therefore <math>x = \dfrac{1}{1-m}</math>.  
 
Solving the systems of equations, we find that <math>mx+3 = (2m-1)x+4</math>, which simplifies to <math>(m-1)x+1 = 0</math>. Therefore <math>x = \dfrac{1}{1-m}</math>.  
 
<math>x</math> is only a real number if <math>\boxed{\textbf{(D) }m\neq 1}</math>.
 
<math>x</math> is only a real number if <math>\boxed{\textbf{(D) }m\neq 1}</math>.
 +
 +
==See Also==
 +
{{AHSME box|year=1975|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Latest revision as of 15:50, 19 January 2021

Problem

For which real values of m are the simultaneous equations

\begin{align*}y &= mx + 3 \\  y& = (2m - 1)x + 4\end{align*}

satisfied by at least one pair of real numbers $(x,y)$?

$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\  \text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m$


Solution

Solution by e_power_pi_times_i


Solving the systems of equations, we find that $mx+3 = (2m-1)x+4$, which simplifies to $(m-1)x+1 = 0$. Therefore $x = \dfrac{1}{1-m}$. $x$ is only a real number if $\boxed{\textbf{(D) }m\neq 1}$.

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS