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Difference between revisions of "1957 AHSME Problems/Problem 41"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(D) }\frac{\pm \sqrt2}2}</math>.
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Notice that the two equations are lines in [[standard form]], so the system will have no solutions if these lines never intersect. For them never to intersect, they must have the same slope and not be the exact same line. We are sure that they are not the same line, because they have different coefficients and have the same number, <math>1</math>, on their right hand sides. In the case that one of the lines has an undefined slope (where it is of the form <math>x=k</math>, where <math>k</math> is some constant), then <math>a</math> is either <math>1</math> or <math>0</math>. In either case, the other line does not have an undefined slope, so the lines eventually intersect and there must be a solution to the system of equations. Thus, we need not worry about this case. Because the lines are in standard form, we can equate the two expressions for their slope as follows:
 +
\begin{align*}
 +
-\frac{a-1}{a} &= - \frac{-a}{a+1} \\
 +
-(a^2-1) &= a^2 \\
 +
2a^2 &= 1 \\
 +
a^2 &= \frac 1 2 \\
 +
a &= \pm \frac 1{\sqrt2} = \pm \frac{\sqrt2}2
 +
\end{align*}
 +
Thus, our answer is <math>\boxed{\textbf{(D) }\frac{\pm \sqrt2}2}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:09, 27 July 2024

Problem

Given the system of equations \[ax + (a - 1)y = 1\] \[(a + 1)x - ay = 1\] For which one of the following values of $a$ is there no solution $x$ and $y$?

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

Solution

Notice that the two equations are lines in standard form, so the system will have no solutions if these lines never intersect. For them never to intersect, they must have the same slope and not be the exact same line. We are sure that they are not the same line, because they have different coefficients and have the same number, $1$, on their right hand sides. In the case that one of the lines has an undefined slope (where it is of the form $x=k$, where $k$ is some constant), then $a$ is either $1$ or $0$. In either case, the other line does not have an undefined slope, so the lines eventually intersect and there must be a solution to the system of equations. Thus, we need not worry about this case. Because the lines are in standard form, we can equate the two expressions for their slope as follows: \begin{align*} -\frac{a-1}{a} &= - \frac{-a}{a+1} \\ -(a^2-1) &= a^2 \\ 2a^2 &= 1 \\ a^2 &= \frac 1 2 \\ a &= \pm \frac 1{\sqrt2} = \pm \frac{\sqrt2}2 \end{align*} Thus, our answer is $\boxed{\textbf{(D) }\frac{\pm \sqrt2}2}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
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

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