Difference between revisions of "1973 AHSME Problems/Problem 10"

Latest revision as of 12:58, 20 February 2020

Problem

If $n$ is a real number, then the simultaneous system

$nx+y = 1$

$ny+z = 1$

$x+nz = 1$

has no solution if and only if $n$ is equal to

$\textbf{(A)}\ -1\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 0\text{ or }1\qquad\textbf{(E)}\ \frac{1}2$

Solutions

Solution 1

Add up all three equations to get $(n+1)(x+y+z) = 3$ If $n = -1$ , then equation results in $0 = 3$ . That has no solutions, so the answer is $\boxed{\textbf{(A)}}$ .

Solution 2

From the first equation, $y = 1 - nx$ Substitute that in the second equation to get $n(1-nx) + z = 1$ $z = 1 - n + n^2 x$ Substitute that in the third equation to get $x + n(1-n+n^2 x) = 1$ $(1+n^3)x + n - n^2 = 1$ If $n = -1$ , then equation results in $-2 = 1$ . That has no solutions, so the answer is $\boxed{\textbf{(A)}}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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
All AHSME Problems and Solutions

@@ Line 28: / Line 28: @@
 <cmath>z = 1 - n + n^2 x</cmath>
 Substitute that in the third equation to get
-<cmath>x + n(1-n+n^2 x) = 1
+<cmath>x + n(1-n+n^2 x) = 1</cmath>
-</cmath>(1+n^3)x + n - n^2 = 1<math></math>
+<cmath>(1+n^3)x + n - n^2 = 1</cmath>
 If <math>n = -1</math>, then equation results in <math>-2 = 1</math>.  That has no solutions, so the answer is <math>\boxed{\textbf{(A)}}</math>.
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=9|num-a=11}}
+{{AHSME 30p box|year=1973|num-b=9|num-a=11}}
 [[Category:Introductory Algebra Problems]]

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