1973 AHSME Problems/Problem 10

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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 $\boxed{\textbf{(A)}}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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