Question

How do you solve `\frac { 4x } { 2x - 1} + \frac { 3} { 2x + 1} \leq 0`?

Answer 1

The solution is `x in [-3/2, -1/2) uu [1/4, 1/2)`

Explanation:

We cannot do crossing over.

Let's rewrite and simplify the inequality

`(4x)/(2x-1)+(3)/(2x+1)<=0`

`(4x(2x+1)+3(2x-1))/((2x-1)(2x+1))<=0`

`(8x^2+4x+6x-3)/((2x-1)(2x+1))<=0`

`(8x^2+10x-3)/((2x-1)(2x+1))<=0`

`((4x-1)(2x+3))/((2x-1)(2x+1))<=0`

Let `f(x)=((4x-1)(2x+3))/((2x-1)(2x+1))`

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3/2` `color(white)(aaaa)` `-1/2` `color(white)(aaaa)` `1/4` `color(white)(aaaaa)` `1/2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `2x+3` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2x+1` `color(white)(aaaa)` `-` `color(white)(aaa)` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `4x-1` `color(white)(aaaa)` `-` `color(white)(aaa)` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `-` `color(white)(a)` `0` `color(white)(a)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2x-1` `color(white)(aaaa)` `-` `color(white)(aaa)` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `-` `color(white)(a)` `color(white)(aa)` `-` `color(white)(a)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `0` `color(white)(aaa)` `-` `color(white)(a)` `||` `color(white)(aa)` `+` `color(white)(a)` `0` `color(white)(a)` `-` `color(white)(a)` `||` `color(white)(aa)` `+`

Therefore,

`f(x)<=0` when `x in [-3/2, -1/2) uu [1/4, 1/2)`

graph{(4x)/(2x-1)+3/(2x+1) [-11.26, 11.25, -5.64, 5.61]}

Answer 2

`[-3/2,-1/2)uu[1/4,1/2)`

Explanation:

`"express the left side as a single fraction"`

`(4x(2x+1))/((2x-1)(2x+1))+(3(2x-1))/((2x-1)(2x+1))<=0`

`rArr(8x^2+10x-3)/((2x-1)(2x+1))<=0`

`"factorise the numerator using the "color(blue)"quadratic formula"`

`rArr((x+3/2)(x-1/4))/((2x-1)(2x+1))<=0`

`"the zeros of the numerator/denominator are"`

`"numerator "x=-3/2,x=1/4,"denominator "x=+-1/2`

`"note that "x!=+-1/2`

`"as this would result in the function being undefined"`

`"the domain is split into 5 intervals as follows"`

`(1)color(white)(x)x<=-3/2color(white)(x)(2)color(white)(x)-3/2<=x<-1/2`

`(3)color(white)(x)-1/2 < x<=1/4color(white)(x)(4)color(white)(x)1/4<=x<1/2color(white)(x)(5)color(white)(x)x>1/2`

`"consider a "color(blue)"test point in each interval"`

`"we want to find where the function is negative "x<=0`

`"substitute each test point into the function and consider"`
`"it's sign"`

`(1)color(white)(x)color(red)(x=-2 )to(+)/(+)tocolor(red)" positive"`

`(2)color(white)(x)color(red)(x=-1)to(-)/(+)tocolor(blue)" negative"`

`(3)color(white)(x)color(red)(x=0)to(-)/(-)tocolor(red)" positive"`

`(4)color(white)(x)color(red)(x=3/8)to(+)/(-)tocolor(blue)" negative"`

`(5)color(white)(x)color(red)(x=1)to(+)/(+)tocolor(red)" positive"`

`"hence the intervals that satisfy the inequality are"`

`-3/2<=x<-1/2" or "1/4<=x<1/2`

`"in interval notation " [-3/2,-1/2)uu[1/4,1/2)`
graph{(8x^2+10x-3)/((2x-1)(2x+1)) [-10, 10, -5, 5]}