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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

njkidd
2005-10-09 11:05:53

Thanks.

I was overdoing the #3.

I got it now.

mathsyperson
2005-10-08 18:58:44

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.

njkidd
2005-10-08 12:36:43

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.

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