I don't know how to properly type up all the mathematical symbols on here so this might look a bit stupid...
-4 > x-4
3x-4 1
But yeah, I can solve this up to where I get
-4 > 3x^2-16x+16
And now what do I do with it??
Add 4 to both sides and then factor the quadratic. After that, put the values that solve the quadratic on a number line and see which ranges make the equation true.
Also, it maybe more in your favor to have a better name than MathSucks123 when wanting help from people who think the contrary.
First of all we have to assume, $\displaystyle x$ cannot be $\displaystyle \frac{4}{3}$ since if it was, it makes the L.H.S. of the orginial inequality undefined.
Multiplying both sides by $\displaystyle (3x-4)^2$, (and since we are sure this is greater than 0), we obtain
$\displaystyle -4(3x-4) \geq (x-4)(3x-4)^2 \Leftrightarrow$
$\displaystyle 0 \geq 4(3x-4)+(x-4)(3x-4)^2\Leftrightarrow$
$\displaystyle 0 \geq \big(4+(x-4)(3x-4)\big)(3x-4)\Leftrightarrow$
$\displaystyle 0 \geq (3x^2-16x+20)(3x-4)\Leftrightarrow$
$\displaystyle 0 \geq (3x-10)(x-2)(3x-4)$
which gives
$\displaystyle x<\frac{4}{3} $ $\displaystyle \big($ strong inequaltiy since $\displaystyle x\ne\frac{4}{3}\big) }$ or $\displaystyle 2\leq x \leq \frac{10}{3_}{.}$
You can actually consider different cases like Soroban did below, but it's up to you which approach you employ.
Hello, MathSucks123!
$\displaystyle \dfrac{\text{-}4}{3x-4} \:\ge\:\dfrac{x-4}{1}$
Do NOT multiply through by an expression containing $\displaystyle \,x$
. . unless you take responsibility for what happens
. . when that expression happens to be negative.
Solve the equation: .$\displaystyle \dfrac{\text{-}4}{3x-4} \:=\:\dfrac{x-4}{1}$
We get: .$\displaystyle 3x^2 - 16x + 20 \:=\:0 \quad\Rightarrow\quad (x-2)(3x-10) \:=\:0$
. . $\displaystyle x \:=\:2,\,\frac{10}{3}$
Note that the first fraction is undefined when $\displaystyle x = \frac{4}{3}$
These divide the number line into four intervals.
. . $\displaystyle \begin{array}{ccccccc}--- & \circ & -- & \bullet & -- & \bullet & --- \\ & 1\frac{1}{3} && 2 && 3\frac{1}{3}\end{array}$
Test a value in each interval and see if it satisfies the inequality.
$\displaystyle x=0\!:\;\;\frac{-4}{0-4} \,\ge\,\frac{0-4}{1} \quad\Rightarrow\quad 1 \,\ge\,-4$ . . . Yes
$\displaystyle x = \frac{3}{2}\!:\;\;\frac{-4}{\frac{9}{2}-4} \,\ge\,\frac{\frac{3}{2}-4}{1} \quad\Rightarrow\quad -8 \,\ge\,-\frac{5}{2}$ . . . no
$\displaystyle x = 3\!:\;\;\frac{-4}{9-4}\,\ge\,\frac{3-4}{1} \quad\Rightarrow\quad -\frac{4}{5} \,\ge\,-1$ . . . Yes
$\displaystyle x = 4\!:\;\;\frac{-4}{12-4} \,\ge\,\frac{4-4}{1} \quad\Rightarrow\quad -\frac{1}{2}\,\ge\,0$ . . no
The solution is: .$\displaystyle \left(-\infty,\:\frac{4}{3}\right)\,\cup\,\left[2,\:\frac{10}{3}\right] $