Question 1210612
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It is not at all clear what the purpose of this problem is.  You can find numerous solutions either by pure trial and error or using formal algebra.<br>
And, as another tutor pointed out, it is not clear whether numbers from the given list can be used more than once.<br>
Using A, B, C, D, and E to represent the numbers in the blanks, the given equation is<br>
<br>
{{{sqrt(2x-3)+Ax+B=Cx-D+E}}}<br>
This equation can be "linear in disguise" if x is any number for which {{{2x-3}}} is a perfect square.<br>
Even limiting ourselves to integer perfect squares, there are an infinite number of possible values for x:<pre>

    2x-3   x
   ----------
     0    3/2
     1      2
     9      6
    25     14
    49     26
    ...</pre>
For many of those values of x the equation will have solutions (often multiple solutions) with A, B, C, D, and E being values from the given list.<br>
But {{{2x-3}}} can be a perfect square which is not an integer; that will lead us to a whole new family of possible values for x, and possibly many of those will allow additional solutions to the problem.<br>
Finally {{{2x-3}}} does not have to be a perfect square.  We can isolate the radical and square both sides of the equation to get yet another whole family of possible solutions to the problem.<br>
{{{sqrt(2x-3)=Cx-D+E-Ax-B}}}
{{{sqrt(2x-3)=(C-A)x+(-D+E-B)}}}<br>
Squaring both sides will yield a quadratic equation; since we are supposed to get a linear equation, it would be necessary for {{{C-A}}} to be zero -- i.e., A and C would have to be the same value from the given list.  If that is indeed allowed, then we open up the possibility of many more solutions to the problem.<br>