Question

Find the nonpermissable replacement for the variable in this expression.
3x/5x+2

Answer 1

can you solve 5x + 2 = 0, what would that give you for "x" ?

Answer 2

-2/5

Answer 3

so let's use that value for "x", let's see what we end up \(\bf \cfrac{3{\color{blue}{ x}}}{5{\color{blue}{ x}}+2}\qquad {\color{blue}{ 5x+2=0\implies x=-\cfrac{2}{5}}}
\\ \quad \\
\cfrac{3{\color{blue}{-\frac{2}{5} }}}{5{\color{blue}{-\frac{2}{5} }}+2}\implies \cfrac{-\frac{6}{5}}{-2+2}\implies \cfrac{-\frac{6}{5}}{{\color{red}{ 0}}}\)

Answer 4

so if we use -2/5 for "x"
our denominator ends up with a big fat 0
and thus our fraction becomes UNDEFINED

thus -2/5 is NOT-permissable for "x", meaning, for our fraction not to become undefined \(\bf x\ne -\cfrac{2}{5}\)

Answer 5

What is the solution set of x for the given equation? x^2/3 - x^1/3 +4=6

Answer 6

\(\large{ \bf x^{\frac{2}{3}}-x^{\frac{1}{3}}+4=6\implies x^{\frac{2}{3}}-x^{\frac{1}{3}}-2=0
\\ \quad \\
\implies ({\color{blue}{ x^{\frac{1}{3}}}})^2-{\color{blue}{ x^{\frac{1}{3}}}}-2=0\qquad let\quad {\color{blue}{ u=x^{\frac{1}{3}}}}\quad \\ \quad \\thus\implies {\color{blue}{ u}}^2-{\color{blue}{ u}}-2=0}\)