SOLUTION: B is taller than j and 3 pillars. P is shorter than B and 2 pillars is j shorter/taller than P?

Question 384383: B is taller than j and 3 pillars. P is shorter than B and 2 pillars is j shorter/taller than P?
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
B is taller than j and 3 pillars. P is shorter than B and 2 pillars is j shorter/taller than P?
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Use "R" to indicate the relation of B to P.
B > j + 3p
p < B + 2p
---
Rearrange:
j < B -3p
j < (B +2p) - 5p
j < P - 5p
j - P < -5p
P-j > 5p
----
Assuming that a pillar (p) is positive,
P is greater than j.
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Cheers,
Stan H.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Let > imply taller
Let < imply shorter

the first statement says that:

B > j + 3

This means that there is a positive value we can add to j + 3 to make it equal to B.

We'll call that value x.

We get:

B = j + 3 + x

The second statement says that:

P < B + 2

We can replace B with j + 3 + x to make that statement become:

P < j + 3 + x + 2

We can simplify this statement to be equal to:

P < j + x + 5

This means that there is a value we can add to P to make it equal to j + x + 5.

We'll call that value y.

The statement becomes:

P + y = j + x + 5

If y = x + 5, then the statement will become:

P + x + 5 = j + x + 5

If we subtract (x + 5) from both sides of the equation, we will then get:

P = j

So, if y = x + 5, then we get P = j.

So P = j is possible when y = x + 5.

Going back to the original equation, we have:

P + y = j + x + 5

If we let y be greater than x + 5, then we get another value (call it z), such that:

y = x + 5 + z

Our equation of P + y = j + x + 5 becomes:

P + x + 5 + z = j + x + 5

If we subtract (x + 5) from both sides of this equation, we get:

P + z = j

This implies that P < j

So we can have P = j and we can have P < j

Going back to the original equation again, we have:

P + y = j + x + 5

If we let y be smaller than x + 5, then we get another value (call it w), such that:

y + w = x + 5 which becomes y = x + 5 - w after we subtract w from both sides of the equation.

Our equation becomes:

P + x + 5 - w = j + x + 5

If we subtract x + 5 from both sides of this equation, we get:

P - w = j

If we add w to both sides of this equation, we get:

P = j + w

This implies that P > j

So, we can get all 3 conditions, depending on the relationship between x and y.

They are:

P = j when y = x + 5
P < j when y > x + 5
P > j when y < x + 5

We can put this into numbers in order to confirm that what we have determined algebraically is correct.

Our starting statements are:

B > j + 3
P < B + 2

We translated these into:

B = j + 3 + x (x > 0)
P + y = B + 2 (y > 0)

Let's let x = 5 and y = 10 (y = x + 5)

We get:

B = j + 3 + 5 which becomes B = j + 8
P + 10 = B + 2 which becomes P = B - 8

Substitute j + 8 for B and we get P = j + 8 - 8 which becomes P = j.

When y = x + 5, we get P = j.

Back to the original equations:

B > j + 3
P < B + 2

We translated these into:

B = j + 3 + x (x > 0)
P + y = B + 2 (y > 0)

Let's let x = 5 and y = 11 (y > x + 5)

We get:

B = j + 3 + 5 which becomes B = j + 8
P + 11 = B + 2

Substitute j + 8 for B in the second equation and we get:

P + 11 = j + 8 + 2 which becomes:
P + 11 = j + 10
Subtract 10 from both sides of the equation to get:
P + 1 = j
This implies that P is less than j (P < j).

Back to the original equations:

B > j + 3
P < B + 2

We translated these into:

B = j + 3 + x (x > 0)
P + y = B + 2 (y > 0)

Let's let x = 5 and y = 9 (y < x + 5)

We get:

B = j + 3 + 5 which becomes B = j + 8
P + 9 = B + 2

Substitute for B in the second equation to get:

P + 9 = j + 8 + 2 which becomes:
P + 9 = j + 10

Subtract 9 from both sides of the equation to get:

P = j + 1

This implies that P > j

Bottom line:

P can be taller than or equal to or shorter than j.

This all depends on the relationship between x and y.

In the translation of the equation B > j + 3 to B = j + 3 + x, x is the amount that B is greater than (taller than) j + 3.

In the translation of the equation P < B + 2, to P + y = B + 2, y is the amount that P is smaller than (shorter than) B + 2.

Making the inequality an equality allows the problem to be solved.

The creation of the difference variables of x, y, z, w allowed the equality to be modeled from the inequality.

x,y,z,w are all assumed to be greater than 0.

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