Difference between revisions of "1975 AHSME Problems/Problem 2"
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+ | ==Problem== | ||
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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> | ||
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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>. | 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
satisfied by at least one pair of real numbers ?
Solution
Solution by e_power_pi_times_i
Solving the systems of equations, we find that , which simplifies to . Therefore .
is only a real number if .
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
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.