# Thread: need help with Solving inequalities using interval notation and grapghing.

1. ## need help with Solving inequalities using interval notation and grapghing.

2. 2x-3/x+1 ≤ 1 (inequality involving quotient)

2. Originally Posted by rsultana

2. 2x-3/x+1 ≤ 1 (inequality involving quotient)
I'll show you how to do the first. The second is done the same way.

$x(x - 1)(x + 2) > 0$

First we want to get everything onto one side of the equation. That's already done here.

Next we want to identify "critical points." This is where the LHS of the equation is 0, or where the denominator is 0. Here we have no denominator, so our critical points are x = -2, 0, 1.

Now we want to break the real line into pieces, given by the critical points. In this case we have 4 intervals:
$( - \infty, -2), (-2, 0), (0, 1), (1, \infty)$

Now we want to test values of x from each interval in the inequality to see which ones give the correct result:
$( - \infty, -2)$ gives $x(x - 1)(x + 2) < 0$ (No!)

$(-2, 0)$ gives $x(x - 1)(x + 2) > 0$ (Yes!)

$(0, 1)$ gives $x(x - 1)(x + 2) < 0$ (No!)

$(1, \infty )$ gives $x(x - 1)(x + 2) > 0$ (Yes!)

Only two of these intervals have x values that work in the inequality. These intervals are the solution set. Thus the answer is $( -2, 0) \cup (1, \infty )$.

The answer to the second problem is $(-1, 4]$. (The 4 is included in the solution set because we have a $\leq$ symbol. Why isn't -1 included in the solution set?)

-Dan

3. Originally Posted by topsquark
I'll show you how to do the first. The second is done the same way.

$x(x - 1)(x + 2) > 0$

First we want to get everything onto one side of the equation. That's already done here.

Next we want to identify "critical points." This is where the LHS of the equation is 0, or where the denominator is 0. Here we have no denominator, so our critical points are x = -2, 0, 1.

Now we want to break the real line into pieces, given by the critical points. In this case we have 4 intervals:
$( - \infty, -2), (-2, 0), (0, 1), (1, \infty)$

Now we want to test values of x from each interval in the inequality to see which ones give the correct result:
$( - \infty, -2)$ gives $x(x - 1)(x + 2) < 0$ (No!)

$(-2, 0)$ gives $x(x - 1)(x + 2) > 0$ (Yes!)

$(0, 1)$ gives $x(x - 1)(x + 2) < 0$ (No!)

$(1, \infty )$ gives $x(x - 1)(x + 2) > 0$ (Yes!)

Only two of these intervals have x values that work in the inequality. These intervals are the solution set. Thus the answer is $( -2, 0) \cup (1, \infty )$.

The answer to the second problem is $(-1, 4]$. (The 4 is included in the solution set because we have a $\leq$ symbol. Why isn't -1 included in the solution set?)

-Dan
i am sorry but i dont get the graphing part of the inequalities.for the second problem i can get to x-4/x+1≤0 but from there on wards i dont know what to do!and i have no idea why -1 is not included.is it because we got the -1 from subtracting it frm the other side?

4. Originally Posted by rsultana
i am sorry but i dont get the graphing part of the inequalities.for the second problem i can get to x-4/x+1≤0 but from there on wards i dont know what to do!and i have no idea why -1 is not included.is it because we got the -1 from subtracting it frm the other side?
Well, I didn't help you with the graphing, all I gave you was what intervals of the real line satisfy the inequality.

The way I learned how to graph the first problem (there are at least two conventions being used)
$( -2, 0) \cup (1, \infty )$
would be to put an open circle on x = -2, x = 0, and x = 1, then draw a line connecting x = -2 and x = 0 to indicate that these points are not part of the solution set, then draw a line to the right starting at x = 1 and continuing to $\infty$. (I put an arrow on the line from x = 1.)

For the second problem:
$(-1, 4]$
I put an open circle on -1 and a closed circle on 4 (to indicate that x = 4 is part of the solution) and connect x = -1 and x = 4.

Why can't x = -1 in the second solution? First of all, PLEASE use parenthesis!!
$x-4/x+1 = x - (4/x) + 1$

What you wanted was "(x - 4)/(x + 1)"

Anyway, plug x = -1 into the original inequality. Is this possible? What about x = 4?

-Dan

5. Originally Posted by topsquark
Well, I didn't help you with the graphing, all I gave you was what intervals of the real line satisfy the inequality.

The way I learned how to graph the first problem (there are at least two conventions being used)
$( -2, 0) \cup (1, \infty )$
would be to put an open circle on x = -2, x = 0, and x = 1, then draw a line connecting x = -2 and x = 0 to indicate that these points are not part of the solution set, then draw a line to the right starting at x = 1 and continuing to $\infty$. (I put an arrow on the line from x = 1.)

For the second problem:
$(-1, 4]$
I put an open circle on -1 and a closed circle on 4 (to indicate that x = 4 is part of the solution) and connect x = -1 and x = 4.

Why can't x = -1 in the second solution? First of all, PLEASE use parenthesis!!
$x-4/x+1 = x - (4/x) + 1$

What you wanted was "(x - 4)/(x + 1)"

Anyway, plug x = -1 into the original inequality. Is this possible? What about x = 4?

-Dan
if we plug in x=-1 then we can get an undefined problem but we get 0 if we plug in x=4