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Neko [114]
4 years ago
5

Graph: 4p+1>-7 or 6p+3<33

Mathematics
1 answer:
Hoochie [10]4 years ago
3 0
This problem involves only one variable, so we stick to one horizontal line, which represents p values.  There is no vertical axis.

If 4p+1>-7, we solve for p by subtracting 1 from both sides:  4p>-8; then we divide both sides by 4, obtaining p>-2  Draw an open circle at p=-2 and from this open circle draw an arrow to the right.

If 6p+3<33, 6p<30.  Dividing both sides by 6, p<5.  Draw an open circle at p=5 and from this open circle draw an arrow to the right.

Now determine the p values for which your two arrows coincide.  The first arrow begins at p=-2 and extends to the right from there; the second arrow begins at p=5 and extends to the left.  So, the only coincidence of the two arrows is between -2 and +5 (noting that the arrows do NOT touch p=-2 or p=5).

The solution set can be writtten as -2<p<5, or as (-2,5).
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