Difference between revisions of "1950 AHSME Problems/Problem 14"
m (→Problem) |
|||
Line 1: | Line 1: | ||
== Problem== | == Problem== | ||
− | For the simultaneous equations <cmath>2x-3y=8 | + | For the simultaneous equations <cmath>2x-3y=8</cmath> |
+ | <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> |
Latest revision as of 16:15, 21 December 2015
Problem
For the simultaneous equations
Solution
Try to solve this system of equations using the elimination method.
Something is clearly contradictory so
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 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.