What number must be placed in the box in the equation below to produce an equation that has more than one solution:
$1/2*y + 1/4 = 3 + \boxed{\phantom{400} } y$
+76 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.
