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**njkidd****Guest**

Please help me.

1. |3x+26|-6<_x

2. Write an absolute value inequality that has the solution all real numbers. Explain. I know that the answer is |t+3|>-3, but i want the explain.

3. Find the value of k so that equation has the solution of -8. |x-2|=k. - I made x as -8, but it doesnt work out well.

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

1. |3x+26|-6 < -x

Sketching the graphs shows that the two lines cross once when |3x+26| is 3x+26, and again when it is -(3x+26).

(1) 3x+26-6 < -x

4x+20 < 0

x < -5

(2) -(3x+26)-6 < -x

3x+26+6 > x

2x > -32

x > -16

Combining these inequalities together gives -16 < x < -5

2. |t+3| > -3 is one of many examples that you could have had. As the whole left hand side has is an absolute value, it cannot be less than 0, so if the right hand side is less than 0, then all values of t will be a solution.

The general solution would be |{Any function of t that can't return undefined or imaginary values}| > {Any negative constant}

3. You need to find a value for k so that when you solve the equation, x will equal -8.

To do this, just substitute in x=-8 and solve for k.

|-8-2|=k

|-10|=k

10=k

Hope I helped.

Why did the vector cross the road?

It wanted to be normal.

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**njkidd****Guest**

Thanks.

I was overdoing the #3.

I got it now.

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