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# If ‘a’ and ‘b’ are real numbers, for what values does the equation $3x-5+a=bx+1$ have a unique solution ‘x’?(a) For all ‘a’ and ‘b’(b) For no ‘a’ and ‘b’(c) For $a\ne 6$ (d) For $b \ne 3$.

Last updated date: 12th Sep 2024
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Hint: Start by rearranging the equation given in the question such that LHS consists of the terms consisting x and all the other terms are in the RHS of the equation. Once, you have simplified the equation, figure out the constraints, and report the answer.

$3x-5+a=bx+1$
$3x-bx=1+5-a$
$x\left( 3-b \right)=6-a$
Now, if we observe, we will find that it is a=6 and b=3, then RHS=LHS=0 for any value of x, so the equation will have infinite solutions in this case. So, we can see that if $b=3$, then x gets diminished and we will not be able to determine the value of “x” and “a” can take as many possible values from the set of real numbers. So, for $b\ne 3$, $3x-5+a=bx+1$ have a unique solution.
Note: The one mistake that a student makes very often is that they solve the equation completely giving $x=\dfrac{6-a}{3-b}$, looking at which we cannot figure out any constraint on a and b and hence reporting the answer as an option (a), which is wrong.