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Linear Algebra. 1. Find the point of intersection:. Let Supply: Q s =3p and Demand: Q d =200-p Q s =3p=3*50=150 Q d =200-p=200-150=150 Equilibrium: Q s =Q d 3p=200-p 4p=200 p=50 Thus Q=3x50=150. 200. S. 50. D. 150. 200. Linear Algebra.

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linear algebra
Linear Algebra

1. Find the point of intersection:

Let Supply: Qs=3p

and Demand: Qd=200-p

Qs=3p=3*50=150

Qd=200-p=200-150=150

Equilibrium: Qs=Qd

3p=200-p

4p=200

p=50

Thus Q=3x50=150

200

S

50

D

150

200

linear algebra2
Linear Algebra

2. Find line across two points: (x,y) = (5,10) and (x,y) = (25,20)

Y=mX+b

10 = m5 + b

20 = m25 + b

b = 10 - 5m

20 = 25m + (10-5m)

20 = 20m + 10

10 = 20m

m = ½

b = 10 - 5m

b = 10 - 5(1/2)

b = 7.5

Therefore… P = ½Q + 7.5

or Q = 2P - 15

20

10

5

25