# B with the 20unit increase in p y the demand curve

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(b) With the \$20/unit increase in P Y , the demand curve becomes X D = 60 – P X or P X = 60 – X D . (c) With the \$2000 increase in M , the demand curve also becomes X D = 60 – P X , or P X = 60 – X D. That is, in this case a \$20/unit increase in P Y has the same effect on X D as a \$2000 increase in M . (d) Note that in this case, the new demand curves are parallel to the original one: the effect of the changes in P Y and M on the demand curve is registered solely in the vertical intercept. This need not be the case, as Answer 6 makes clear. M1-5 MATH MODULE 1: SOLUTIONS TO EXERCISES P X 0 a b = c (50, 0) (60, 0) (0, 60) (0, 50) X D FIGURE FOR ANSWER M.1-5

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6. In this case, because of the “interaction term” P X P Y , the change in P Y alters both the vertical and horizontal intercepts and the slope. Note also that in both question 5 and this one, X is a normal good (since the coefficient of the M term is positive), but that X and Y are substitutes in question 5 (since there an increase in P Y increases the quan- tity demanded of X at any P X ), while in this question they are complements (since here an increase in P Y would decrease the quantity demanded of X at any P X ). (a) Substituting into X D = 38 – P X – 0.05 P X P Y – 0.4 P Y + 2 M , we get X D = 38 – P X 0.05 P X (20) – 0.4(20) + 2(10) = 50 – 2 P X , or P X = 25 – 0.5 X D. (b) With the decrease in the price of the complement Y, demand becomes X D = 54 – 1.5 P X , or P X = 36 – (2/3) X D . M1-6 MATH MODULE 1: SOLUTIONS TO EXERCISES P X 36 Slope = 2 25 0 50 54 X D 3 Slope = 1 2 FIGURE FOR ANSWER M.1-6
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