If p+q = 12, q+r= 24, P+R=24 then p*q*r= ?

Question

If P,Q and R are integers satisfying the equations
p+q = 12,
q+r= 24,
P+R=24
then what is their product ? (p*q*r= ?)

Answer 1

(6)(6)(18) = ...

Answer 2

I'll take the first two equations and set them to be q equal to everything else.



p + q = 12

q = 12 - p



q + r = 24

q = 24 - r



Now that I have two equations equal to q, the other halves must also be equal to each other.



12 - p = 24 - r



that and the third equation :



p + r = 24



makes a system of two equations and two unknowns. set the "second" to p in terms of r and substitute it into the "first", then solve for r.



p = 24 - r



12 - p = 24 - r

12 - (24 - r) = 24 - r

12 - 24 + r = 24 - r

-12 + r = 24 - r

2r = 36

r = 18



Now we have r, we can solve for q and p by plugging the value of r into the second and third of the original equations.



q + r = 24

q + 18 = 24

q = 6



p + r = 24

p + 18 = 24

p = 6



Now that we know what p, q, and r are ...



pqr = 6(6)(18) = 648

Answer 3

12 + 24 - 24 = 12

2q = 12

q = 6

p = 6

r = 18



pqr = 18 * 36 = 648

Answer 4

q + r = 24

p + q = 12



SUBTRACT



r - p = 12

p+r = 24



2r = 36

r = 18

p = 6

q = 6



pqr = 6*6*18 = 648 ANSWER

Answer 5

p+q = 12

p-q = 0

p = 6

q = 6

r = 18

pqr = 648

Answer 6

I am assuming that the system of equations is



pq = 12

qr = 18

pr = 24.



Multiplying the three equations gives

(pqr)^2 = 12*18*24 = 12*36*12 = (12*6)^2 = 72^2.



Therefore, pqr = +-72.



(Note that for any solution (p1, q1, r1) of the system, (-p1, -q1, -r1) will also be a solution; note that p1*q1*r1 is the negative of (-p1)(-q1)(-r1), so pqr could have either sign.)



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