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

(Created page with "For which real values of m are the simultaneous equations <cmath> \begin{align*}y &= mx + 3 \\ y& = (2m - 1)x + 4\end{align*} </cmath> satisfied by at least one pair of re...")
 
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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>.
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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>.

Revision as of 11:57, 15 December 2016

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 $\boxed{\textbf{(D) }m\neq 1}$.

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