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  1. For what values of $a$ does this equation have no solution in $x$? $$\frac1{\;1+\dfrac1x\;} = a$$
  2. Similarly, for what value of $a$ does this equation have no solution in $x$? $$\frac{6x-a}{x-3} = 3$$

I get only the value $a = 1$ for the first, and $a = 18$ for the second with my method.

I basically delineate which values of $x$ would be unacceptable, such as $x=-1$ for the first equation, and then solve for $x$ in terms of $a$. If that results in an expression with $a$ that calls for excluding a value of $a$ that would result in a denominator of $0$, I exclude that.

I also solve for the value of $a$ that would make $x$ equal a prohibited $x$ value and exclude that.

I'm not sure that's the correct method, so I would appreciate if anyone could run through their solutions. Thanks!

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Sami A
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1 Answers1

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  1. Imagine there is a solution $(a,x)$. If $x\neq0$, the equation is equivalent to $\frac{x}{x+1}=a$. Now if $x\neq-1$, the equation is equivalent to $x=a(x+1)=ax+a$. From which, $$x(1-a)=a\tag{1}$$ And now if $a\neq1$, $$x=\frac{a}{1-a}\tag{2}$$ So as long as $a\neq1$ and also $a$ is not such that equation (2) implies $x=0$ or $x=-1$, there is a solution. Is there a value of $a$ where the equation (2) implies $x=0$? Yes: $a=0$. Is there a value of $a$ where equation (2) implies $x=-1$? No, because equation (1) would be $-1+a=a$, equivalent to $-1=0$.
    So the values of $a$ that lead to no solution are $1$ and $0$.

  2. Imagine there is a solution $(a,x)$. If $x\neq3$, the equation is equivalent to $6x-a=3(x-3)=3x-9$. From which, $$x=\frac{a-9}{3}\tag{3}$$ So as long as $a$ is such that equation (3) does not imply $x=3$, there is a solution. Is there a value of $a$ such that equation (3) implies $x=3$? Yes, $a=18$. There are no other considerations. So $a=18$ is the only value for $a$ for which there is no solution for $x$.

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