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3 Responses
Your setup is on the right track. Write the distances as AP = |P + 3| and BP = |P − 5|. Because P is to the right of B, we know P > 5, so both expressions are already positive. I actually tripped for a moment by writing BP = 5 − P (out of habit), which would give P + 3 = 2(5 − P) and lead to P = 7/3-clearly not to the right of B-so that sign was backward for P > 5. Using the correct BP = P − 5, we get P + 3 = 2(P − 5), hence P = 13. Quick check: AP = 16 and BP = 8, so AP is indeed twice BP. A small analogy: think of a tug-of-war where A pulls twice as hard as B; the “balance point” shifts closer to the weaker team (B), which is why P ends up on B’s side but farther out at 13.
If you ignore the “to the right of B” condition and solve |P + 3| = 2|P − 5| in general, you do get another solution between A and B at P = 7/3; the right-of-B requirement simply rules that one out. I briefly wondered if a different sign choice might produce a third point to the left of A, but going through the absolute-value cases carefully collapses everything to just these two values. If you like a more systematic viewpoint, this is the same as dividing the segment externally in the ratio 2:1; Khan Academy has a nice overview of dividing line segments in a given ratio: https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-dividing-segments/a/dividing-line-segments.
Your setup is spot on! Since P is told to be to the right of B, we know P > 5, which makes both distances positive: AP = |P + 3| becomes P + 3 (because P + 3 > 8) and BP = |P − 5| becomes P − 5. So the equation “P is twice as far from A as from B” really is P + 3 = 2(P − 5). Solving gives P + 3 = 2P − 10, so P = 13. A quick check: distance to A is 16, to B is 8-ta‑da, twice as far! If you didn’t use the “to the right of B” clue, you’d solve |P + 3| = 2|P − 5| and get two solutions: P = 13 and P = 7/3; the second one sits between A and B, so the “to the right” condition neatly tosses it off the number-line stage. For a refresher on turning “distance” into absolute value equations, this guide is handy: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:absolute-value/abs-val-equations/a/solving-absolute-value-equations. As a follow-up puzzle: what point would you get if P were “twice as far from B as from A,” still staying to the right of B?
Your setup is exactly right. Since “P is to the right of B,” we know P > 5, so both distances AP = P + 3 and BP = P − 5 are positive and we can write P + 3 = 2(P − 5). Solving: P + 3 = 2P − 10, so 13 = P. That means P = 13, and a quick check says AP = 16 and BP = 8, so AP is indeed twice BP.
If the problem hadn’t told you “to the right of B,” you’d keep the absolute values and solve |P + 3| = 2|P − 5| piecewise. That gives two algebraic solutions: P = 13 and P = 7/3. The latter sits between A and B (so it gets tossed by the “right of B” condition). Funny aside: it might look like these two are symmetric about the midpoint 1, but that’s just an illusion here. Hope this helps!