Solving Rational Inequalities

**MathSucks123** · November 21st 2010, 01:24 PM

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??

**dwsmith** · November 21st 2010, 01:47 PM

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.

**willy0625** · November 21st 2010, 03:54 PM

First of all we have to assume, $x$ cannot be $\frac{4}{3}$ since if it was, it makes the L.H.S. of the orginial inequality undefined.

Multiplying both sides by $(3x-4)^2$ , (and since we are sure this is greater than 0), we obtain

$-4(3x-4) \geq (x-4)(3x-4)^2 \Leftrightarrow$
$0 \geq 4(3x-4)+(x-4)(3x-4)^2\Leftrightarrow$
$0 \geq \big(4+(x-4)(3x-4)\big)(3x-4)\Leftrightarrow$
$0 \geq (3x^2-16x+20)(3x-4)\Leftrightarrow$
$0 \geq (3x-10)(x-2)(3x-4)$

which gives

$x<\frac{4}{3}$ $\big($ strong inequaltiy since $x\ne\frac{4}{3}\big) }$ or $2\leq x \leq \frac{10}{3_}{.}$

You can actually consider different cases like Soroban did below, but it's up to you which approach you employ.

**Soroban** · November 21st 2010, 03:57 PM

Hello, MathSucks123!

$\dfrac{\text{-}4}{3x-4} \:\ge\:\dfrac{x-4}{1}$

Do NOT multiply through by an expression containing $\,x$
. . unless you take responsibility for what happens
. . when that expression happens to be negative.

Solve the equation: . $\dfrac{\text{-}4}{3x-4} \:=\:\dfrac{x-4}{1}$

We get: . $3x^2 - 16x + 20 \:=\:0 \quad\Rightarrow\quad (x-2)(3x-10) \:=\:0$

. . $x \:=\:2,\,\frac{10}{3}$

Note that the first fraction is undefined when $x = \frac{4}{3}$

These divide the number line into four intervals.

. . $\begin{array}{ccccccc}--- & \circ & -- & \bullet & -- & \bullet & --- \\ & 1\frac{1}{3} && 2 && 3\frac{1}{3}\end{array}$

Test a value in each interval and see if it satisfies the inequality.

$x=0\!:\;\;\frac{-4}{0-4} \,\ge\,\frac{0-4}{1} \quad\Rightarrow\quad 1 \,\ge\,-4$ . . . Yes

$x = \frac{3}{2}\!:\;\;\frac{-4}{\frac{9}{2}-4} \,\ge\,\frac{\frac{3}{2}-4}{1} \quad\Rightarrow\quad -8 \,\ge\,-\frac{5}{2}$ . . . no

$x = 3\!:\;\;\frac{-4}{9-4}\,\ge\,\frac{3-4}{1} \quad\Rightarrow\quad -\frac{4}{5} \,\ge\,-1$ . . . Yes

$x = 4\!:\;\;\frac{-4}{12-4} \,\ge\,\frac{4-4}{1} \quad\Rightarrow\quad -\frac{1}{2}\,\ge\,0$ . . no

The solution is: . $\left(-\infty,\:\frac{4}{3}\right)\,\cup\,\left[2,\:\frac{10}{3}\right]$

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