5. (A version of Blanchard (2017), #5 on pp. 127-128.) Consider the following numerical example of the IS-LM model: C = 300 + 0.4YD I = 180 + 0.2Y - 1000i T = 125 G = 50 MS P = 300 Md P = Y - 5000i (a) Derive the IS relation (HINT: You want an equation with Y on the left side, all else on the right.) (b) Derive the LM relation (HINT: It will be convenient for later use to rewrite this equation with i on the left side, all else on the right.) (c) Solve for equilibrium real output. (HINT: Substitute the expression for the interest rate given by the LM equation into the IS equation, and solve for output.) (d) Solve for the equilibrium interest rate. (HINT: Substitute the value you obtained for Y in part (c) into either the IS equation or the LM equation, and solve for i. If your algebra is correct, you should get the same answer from both equations.) (e) Solve for the equilibrium values of C and I, and verify the value you obtained for Y by adding up C, I, and G