Question

solving inequalities involving rational expressions
learning goals:

• determine solutions to rational inequalities in one variable by identifying intervals for which x satisfies the inequalities
• represent the solutions to a rational inequality using interval notation, number line or algebraic notation

examples:

1. solve the following and identify the restrictions.

a) \\(\frac{x - 2}{x + 3}=\frac{x - 4}{x + 5}\\)
c) \\(\frac{x + 1}{2x - 1}=\frac{x - 4}{3x - 8}\\)

1. solve the following inequality using an interval table. show your solution in interval notation and a number line

a) \\(\frac{2x}{3x + 4}>\frac{x}{x + 1}\\)
b) \\(\frac{x - 2}{x + 1}>\frac{x - 6}{x - 2}\\)

Explanation:

Problem 1a: $\frac{x-2}{x+3}=\frac{x-4}{x+5}$

Step1: Identify restrictions

Denominators cannot be zero:
$x+3
eq0 \implies x
eq-3$
$x+5
eq0 \implies x
eq-5$

Step2: Cross-multiply to eliminate fractions

$(x-2)(x+5)=(x-4)(x+3)$

Step3: Expand both sides

$x^2+5x-2x-10=x^2+3x-4x-12$
$x^2+3x-10=x^2-x-12$

Step4: Simplify to solve for $x$

Subtract $x^2$ from both sides:
$3x-10=-x-12$
$3x+x=-12+10$
$4x=-2$
$x=\frac{-2}{4}=-\frac{1}{2}$

Step1: Identify restrictions

Denominators cannot be zero:
$2x-1
eq0 \implies x
eq\frac{1}{2}$
$3x-8
eq0 \implies x
eq\frac{8}{3}$

Step2: Cross-multiply to eliminate fractions

$(x+1)(3x-8)=(x-4)(2x-1)$

Step3: Expand both sides

$3x^2-8x+3x-8=2x^2-x-8x+4$
$3x^2-5x-8=2x^2-9x+4$

Step4: Simplify to solve for $x$

Subtract $2x^2-9x+4$ from both sides:
$x^2+4x-12=0$

Step5: Factor quadratic equation

$(x+6)(x-2)=0$
$x+6=0$ or $x-2=0$

Step1: Rearrange inequality to 0

$\frac{2x}{3x+4}-\frac{x}{x+1}>0$

Step2: Combine fractions

$\frac{2x(x+1)-x(3x+4)}{(3x+4)(x+1)}>0$

Step3: Simplify numerator

$2x^2+2x-3x^2-4x=-x^2-2x=-x(x+2)$
Inequality becomes: $\frac{-x(x+2)}{(3x+4)(x+1)}>0$
Multiply by -1 (reverse inequality): $\frac{x(x+2)}{(3x+4)(x+1)}<0$

Step4: Find critical points

$x=0, x=-2, x=-\frac{4}{3}, x=-1$ (denominator zeros)

Step5: Test intervals

Interval	$\frac{x(x+2)}{(3x+4)(x+1)}$	Sign
$(-2,-\frac{4}{3})$	$\frac{(-)(+)}{(-)(-)}=-$	Negative
$(-\frac{4}{3},-1)$	$\frac{(-)(+)}{(+)(-)}=+$	Positive
$(-1,0)$	$\frac{(-)(+)}{(+)(+)}=-$	Negative
$(0,\infty)$	$\frac{(+)(+)}{(+)(+)}=+$	Positive

We need intervals where the expression is negative.

Answer:

Restrictions: $x
eq-3, x
eq-5$; Solution: $x=-\frac{1}{2}$

---

Problem 1c: $\frac{x+1}{2x-1}=\frac{x-4}{3x-8}$