solve so that {x|? <= x <= ?}

Hi, Wolfeden!

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Thanks!

In solving inequality problems, the main goal is to isolate X (the unknown variable) on one side.

1. You can add or subtract same number from both sides.

For example 5x-7<= -8
=>5x-7+7<=-8+7 (I want to isolate x)
=>5x<=-1
2. You can divide or multiply both sides by same number

5x/5<=-1/5
=>x<=-1/5
3. Dividing or multiplying both sides by a negative number reverses the inequality

(-1)x>=(-1/5)(-1)
=>-x>=1/5

Now solve your problem part by part-
1. 3x-6<=-1
2. 3x-6>=-17

x<=? X>=? So,{x|___ <= x <= ___}

Hm.. quite hard to get without the LaTeX...

I'll give one more example.

If

5 \leq 2x + 5 \leq 9

You can subtract 5 from all the three sides to give:

5-5 \leq 2x +5-5\leq 9-5

0 \leq 2x \leq 4

Then, you can divide everywhere by 2 to give:

\frac02 \leq \frac{2x}{2} \leq \frac42

0 \leq x \leq 2

So that'll be your answer. If you want to make sure that it's the right answer, try out one value of x (eg. 0, 1, or 2) into the initial inequality. You should get the value between 5 and 9 inclusive.

To go further, say you divide by a negative number... like -1

\frac{0}{-1}\ \leq\ \frac{x}{-1}\ \leq\ \frac{2}{-1}

0 \leq -x \leq -2

Let's see if that is still good... 0 is less than or equal to -x... but you know that from above, x is between 0 and 2, and if you let x = 1, for example, the equation 0 \leq -1 is not true! Conversely, -x \leq -2 is not true either! What happened? Well, whenever you divide or multiply by a negative number, the > and < signs flip over, like this:

0 \geq -x \geq -2

Now, that inequality is valid for the previous solution.