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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 $b \ne 3$.

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Last updated date: 17th Jun 2024
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Answer
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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.

Complete step-by-step answer:
Let us start the solution to the above question by solving the equation given in the question.
$3x-5+a=bx+1$
Now, we will take all the terms consisting of x to one side of the equation, and all the other terms to the other side of the equation. On doing so, we get
$3x-bx=1+5-a$
Now, we will take x common from the terms in the LHS of the equation. On doing so, we get
$x\left( 3-b \right)=6-a$
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.
Hence, The answer to the above question is the option (d).

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.