algebra with fractions

Question

While studying math, I have come across this problem:

$$x < \frac x3 + \frac 23$$

Now according to the textbook, the answer is $x < 1$, but I can't see how they came up with that solution, as I could only come up with the answer $x < x + 2$. So I'd appreciate it if someone could explain this to me.

Answer 1

Subtracting $\frac{x}{3}$ from both sides $$\frac{2x}{3}<\frac23$$ Multiplying both sides by $\frac32$ gives $$x<1$$

Answer 2

$$x<\frac x3+\frac23\stackrel{\cdot3}\implies3x<x+2\stackrel{-x}\implies 2x<2\stackrel{\div2}\implies x<1$$

Observe that all the involved operations, carried on both sides of the inequality, do not change the inequality sign...

Answer 3

The correct answer would be:

\begin{align}x&<\frac x3 +\frac 23\\\\ x&<\frac{x+2}{3}\tag{combine fractions}\\\\ 3x&<x+2\tag{multiply by $3$}\\ 2x&<2\tag{subtract $x$}\\ x&<1\tag{divide by $2$}\end{align}

Your answer is close, it appears that when you multiplied by $3$, you only multiplied the right hand side of the equation, and not also the left, leaving you with $x<x+2$ when you should have had $3x<x+2$

Answer 4

$x-\frac x3 < \frac 23$

$\frac{2x}{3} < \frac 23$

$2x < 2$

$x<1$

Answer 5

\begin{align*} &x < (x/3) + (2/3)\\[4pt] \iff\; &3x < 3{\large{(}}(x/3) + (2/3){\large{)}}\\[4pt] \iff\; &3x < 3(x/3) + 3(2/3)\\[4pt] \iff\; &3x < x + 2\\[4pt] \iff\; &2x < 2\\[4pt] \iff\; &x < 1\\[4pt] \end{align*}