Please check my calculation so far. Thanks

val1

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Oct 17, 2005
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I'm trying to solve this inequality, but first I need to find out what the points of interest
are.
I've got to the point in the solution where I thought I should be able to factorise but the equation I'm getting an awkward set of figures.

Please can you check to see if I've done the calculation correctly so far to save me from going down completely the wrong path with this.
Thanks

The inequality is
\(\displaystyle 30/ (x+1) < x+2\)

Bringing all the terms onto the LHS:
\(\displaystyle 30/ (x+1) - x-2= 0\)
Multiply through by (x+1)
\(\displaystyle [30-(x-2)(x+1)] / (x+1) = 0\)
\(\displaystyle (30-x^2 +x+2)/ (x+1) = 0\)
\(\displaystyle (-x^ +x+32) / (x+1) = 0\)
Multiply top line by –1
\(\displaystyle (x^2 +x+32) / (x+1) = 0\)

At this stage I'm not sure Ive done it right so far because I don't seem to be easily able to factorise this equation.

Can you help me to progress this?

Thanks
 
You can't arbitrarily change "less than" to "equals", and you can't multiply an inequality by an unknown, since you won't know if you have to flip the inequality or not. Also, if you "multiply through" by the denominator, then the denominator should cancel off; it shouldn't still be there.

. . . . .\(\displaystyle \Large{\frac{30}{x\mbox{ }+\mbox{ }1}\mbox{ }<\mbox{ }x\mbox{ }+\mbox{ }2}\)

. . . . .\(\displaystyle \Large{\frac{30}{x\mbox{ }+\mbox{ }1}\mbox{ }-\mbox{ }(x\mbox{ }+\mbox{ }2)\mbox{ }<\mbox{ }0}\)

. . . . .\(\displaystyle \Large{\frac{30}{x\mbox{ }+\mbox{ }1}\mbox{ }-\mbox{ }\frac{(x\mbox{ }+\mbox{ }2)(x\mbox{ }+\mbox{ }1)}{x\mbox{ }+\mbox{ }1}\mbox{ }<\mbox{ }0}\)

. . . . .\(\displaystyle \Large{\frac{30}{x\mbox{ }+\mbox{ }1}\mbox{ }-\mbox{ }\frac{x^2\mbox{ }+\mbox{ }3x\mbox{ }+\mbox{ }2}{x\mbox{ }+\mbox{ }1}\mbox{ }<\mbox{ }0}\)

. . . . .\(\displaystyle \Large{\frac{-(x^2\mbox{ }+\mbox{ }3x\mbox{ }-\mbox{ }28)}{x\mbox{ }+\mbox{ }1}\mbox{ }<\mbox{ }0}\)

Factorize. Solve each factor to find the zeroes of the associated rational function. These zeroes will split the number line into four intervals. Check the sign of the rational on each interval.

The intervals on which the rational is negative are your solution.

Eliz.
 
Thanks for your help

Thanks for replying so promptly.

I had got a similar top line from an earlier attempt, which factorised easily, but when I was going over the solution again I realised I'd missed a negative sign and got into a real mess thrying to sort it out, as you can see!

I really appreciate your help in setting me straight again.

Val
 
Hello, val1!

This type of inequality demands extra care . . .

Recall that important rule for Inequaltieies:
. . When multiplying or dividing by a <u>negative</u> quantity, reverse the inequality.

\(\displaystyle \L\frac{30}{x\,+\,1}\:<\:x\,+\,2\)
First, I will rewrite the statement (for convenience): .\(\displaystyle \L x\,+\,2\:>\:\frac{30}{x\,+\,1}\)

We can eliminate the fraction by multiplying both sides by \(\displaystyle (x\,+\,1)\)
But we must be very careful . . . is \(\displaystyle (x\,+\,1)\) positive or negative?
. . Since we don't know, we must consider both cases.

[1] Suppose \(\displaystyle x\,+\,1\:>\:0\) . . . that is: \(\displaystyle x\,>\,\)-\(\displaystyle 1\) (a)
. . .Then we have: .\(\displaystyle (x\,+\,2)(x\,+\,1)\:>\:30\)
. . . . .which simplifies to: .\(\displaystyle x^2\,+\,3x\,-\,28\:>\:0\)
. . . . .and factors: .\(\displaystyle (x\,-\,4)(x\,+\,7)\:>\:0\)
. . .We find that this is satisfied if: .\(\displaystyle x\,<\,-7\) or \(\displaystyle x\,>\,4\)
. . .But (a) says: \(\displaystyle x\,>\,-1.\) .So the solution set is: \(\displaystyle x\,>\,4\)


[2] Suppose \(\displaystyle x\,+\,1\:<\:0\) . . . that is: \(\displaystyle x\,<\,\)-\(\displaystyle 1\) (b)
. . . Then we are multiplying by a <u>negative</u> quantity.
. . .and we have: .\(\displaystyle (x\,+\,2)(x\,+\,1)\:<\:30\)
. . . . .which simplifies to: .\(\displaystyle x^2\,+\,3x\,-\,28\:<\:0\)
. . . . .and factors: .\(\displaystyle (x\,-\,4)(x\,+\,7)\:<\:0\)
. . .We find that this is satisfied if: .-\(\displaystyle 7\,<\,x\,<\,4\)
. . .But (b) says: \(\displaystyle x\,<\,-1.\) .So the solution set is: -\(\displaystyle 7\,<\,x\,<\,-1\)

Answer: .\(\displaystyle (-7,\,-1)\:\cup\;(4,\,\infty)\)
 
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