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What number must be placed in the box in the equation below to produce an equation that has more than one solution:

$1/2*y + 1/4 = 3 + \boxed{\phantom{400} } y$

 Dec 22, 2021
 #1
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first, re-write the equation in LaTeX.

\({1\over2}y+{1\over4}=3+\boxed{\phantom{400}}y\)

Instead of a box, substitute it for x.

\({1\over2}y+{1\over4}=3+xy\)

To get rid of the fractions, multiply both sides by 4.

\(4*({1\over2}y+{1\over4})=(3+xy)*4\)

simplified...

\(2y+1=12+4xy\)

move all numbers to one side

\((2y+1)-(2y+1)=12+4xy-(2y+1)\)

simplified...

\(4xy-2y+11=0\)

the easiest way to get more than one solution is to make the equation a binomial.

to make \(4xy-2y+11=0\) a binomial, we can just say that x=y

simplified...

\(4y^2-2y+11=0\)

since x=y, y must be placed in the box.

The only thing I am worried about is that it might not want a variable for the box, but instead an actual number.

 Dec 23, 2021

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