Difference between revisions of "1975 AHSME Problems/Problem 2"
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| + | ==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> | ||
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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>. | + | 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>. | ||
| + | |||
| + | ==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. 
