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Y = C + I + G C = 120 + 0.5(Y – T) I = 100 – 10r G = 50 T = 40 L(r,Y) = Y – 20r M = 600 P = 2 a. Using the information above, derive the equation for the IS curve. b. Using the information above, derive the equation for the LM curve. c. What are the equilibrium levels of income and interest rate

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Answer:

a. Y = 500 - 20r

b. Y = 300 + 20r

c. r = 5; Y = 400

Step-by-step explanation:

a. Using the information above, derive the equation for the IS curve.

Y = C + I + G

Substituting for all the values, we have:

Y = 120 + 0.5(Y – 40) + 100 - 10r + 50

Y = 120 + 0.5Y - 20 + 100 - 10r + 50

Y - 0.5Y = 120 - 20 + 100 + 50 - 10r

(1 - 0.5)Y = 250 - 10r

0.5Y = 250 - 10r

Divide through by 0.5, we have:

Y = 500 - 20r <--------- IS curve Equation ............. (1)

b. Using the information above, derive the equation for the LM curve

M/P = L(r,Y)

Substitution for the values, we have:

600/2 = Y - 20r

300 = Y - 20r

Y = 300 + 20r <--------- LM curve Equation ............... (2)

c. What are the equilibrium levels of income and interest rate

Substitute Y in (2) into equation (1), we have:

300 + 20r = 500 - 20r

20r + 20r = 500 - 300

40r = 200

r = 200/40

r = 5 <--------- equilibrium level of interest rate .............. (3)

Substitute for r in equation (2), we have:

Y = 300 + 20(5) = 300 + 100

Y = 400 <--------- equilibrium level of income ................... (4)

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User Tony Park
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