ProblemSet1 Solutions

ProblemSet1 Solutions - f ned Problem 5 If b d = 0 then d =...

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Department of Economics Ryerson University ECN230-Mathematics For Economics Problem Set #1-Solutions EXERCISE 3.2 Problem 1: (a) Q d = Q s 21 3 P = 4+8 P P = 25 11 =2 3 11 . Q =21 3 P Q = 156 11 =14 2 11 . (b) a =21 ,b =3 ,c =4 ,d =8 P = 21 + 4 3+8 = 25 11 =2 3 11 . Q = (21) (8) (3) (4) 3+8 = 156 11 =14 2 11 . Problem 2: (a) Q d = Q s 51 3 P = 10 + 6 P P = 61 9 =6 7 9 . Q =51 3 P Q = 92 3 =30 2 3 . 1
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(b) Q d = Q s 30 2 P = 6+5 P P = 36 7 =5 1 7 . Q =30 2 P Q = 138 7 =19 5 7 . Problem 3: For Problem 1: a =21 ,b =3 ,c =4and d = 8. Therefore, ad bc = 156 > 0and b + d =11 > 0. The condition is veri f ed. For Problem 2(a): a =51 ,b =3 ,c =10and d =6. Therefore, ad bc = 276 > 0and b + d =9 > 0. The condition is veri f ed. For Problem 2(b): a =30 ,b =2 ,c =6and d = 5. Therefore, ad bc = 138 > 0and b + d =7 > 0. The condition is veri f ed. Problem 4: No. If b + d =0 ,then P and Q in expressions (3.4) and (3.5) would
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Unformatted text preview: f ned. Problem 5: If b + d = 0, then d = − b and the demand and supply curves would have the same slope (though di f erent vertical intercepts). The two curves would be parallel to each other, with no equilibrium intersection point as in Figure 3.1. EXERCISE 3.3 Problem 2: (a) ∆ = ( − 8) 2 − 4 (1) (15) = 4 > Hence, there are two real roots: x ∗ 1 = 8 + √ 4 2 (1) = 5 2 and x ∗ 2 = 8 − √ 4 2 (1) = 3. (b) ∆ = ( − 4) 2 − 4 (2) ( − 16) = 144 > Hence, there are two real roots: x ∗ 1 = 4 + √ 144 2 (2) = 4 and x ∗ 2 = 4 − √ 144 2 (2) = − 2. 3...
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ProblemSet1 Solutions - f ned Problem 5 If b d = 0 then d =...

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