Answer:
−5
Explanation:
First, we observe that this is a linear equation.
A linear equation in one variable will have no solutions if the equation reduces to an equation of the form:
\blue a x+\red b=\blue c x + \red dax+b=cx+dstart color #6495ed, a, end color #6495ed, x, plus, start color #df0030, b, end color #df0030, equals, start color #6495ed, c, end color #6495ed, x, plus, start color #df0030, d, end color #df0030
where \blue a=\blue ca=cstart color #6495ed, a, end color #6495ed, equals, start color #6495ed, c, end color #6495ed and \red b\\eq \red db
=dstart color #df0030, b, end color #df0030, does not equal, start color #df0030, d, end color #df0030.
In this case, the equation will reduce to the statement \red b=\red db=dstart color #df0030, b, end color #df0030, equals, start color #df0030, d, end color #df0030, which is not true for any value of xxx.
Hint #2
Since \red {\dfrac23}\\eq \red {\dfrac13}
3
2
=
3
1
start color #df0030, start fraction, 2, divided by, 3, end fraction, end color #df0030, does not equal, start color #df0030, start fraction, 1, divided by, 3, end fraction, end color #df0030 , the equation will have no solutions if \blue b=\blue {-5}b=−5start color #6495ed, b, end color #6495ed, equals, start color #6495ed, minus, 5, end color #6495ed.
Let's check that this is the case. If we add \blue {5}{x}5xstart color #6495ed, 5, end color #6495ed, x to both sides, we get
\begin{aligned} \red{\dfrac23}\blue{-5}x&= \blue{-5} x+\red{\dfrac13}\\\\ \red{\dfrac23}\blue{-5}x+\blue5x&= \blue{-5}x+\blue{-5}x+\red{\dfrac13} \\\\ \red{\dfrac23}&=\red{\dfrac13} \end{aligned}
3
2
−5x
3
2
−5x+5x
3
2
=−5x+
3
1
=−5x+−5x+
3
1
=
3
1
which is not true for any value of xxx, so there are no solutions.
For all other values of b,b,b, comma there will be one solution.
Hint #3
If b=-5b=−5b, equals, minus, 5, the equation will have no solutions.